{"id":1586,"date":"2021-07-22T16:33:33","date_gmt":"2021-07-22T16:33:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1586"},"modified":"2022-03-19T04:03:25","modified_gmt":"2022-03-19T04:03:25","slug":"partial-fractions-with-linear-factors","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/partial-fractions-with-linear-factors\/","title":{"raw":"Partial Fractions With Linear Factors","rendered":"Partial Fractions With Linear Factors"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Integrate a rational function using the method of partial fractions<\/li>\r\n \t<li>Recognize simple linear factors in a rational function<\/li>\r\n \t<li>Recognize repeated linear factors in a rational function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165042047881\">We have seen some techniques that allow us to integrate specific rational functions. For example, we know that<\/p>\r\n\r\n<div id=\"fs-id1165041842105\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{du}{u}=\\text{ln}|u|+C\\text{ and }\\displaystyle\\int \\frac{du}{{u}^{2}+{a}^{2}}=\\frac{1}{a}{\\tan}^{-1}\\left(\\frac{u}{a}\\right)+C\\text{.}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041797885\">However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately obvious how to go about evaluating [latex]\\displaystyle\\int \\frac{3x}{{x}^{2}-x - 2}dx[\/latex]. However, we know from material previously developed that<\/p>\r\n\r\n<div id=\"fs-id1165042089867\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\left(\\frac{1}{x+1}+\\frac{2}{x - 2}\\right)dx=\\text{ln}|x+1|+2\\text{ln}|x - 2|+C[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041972469\">In fact, by getting a common denominator, we see that<\/p>\r\n\r\n<div id=\"fs-id1165040736457\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{x+1}+\\frac{2}{x - 2}=\\frac{3x}{{x}^{2}-x - 2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041913720\">Consequently,<\/p>\r\n\r\n<div id=\"fs-id1165042275619\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{3x}{{x}^{2}-x - 2}dx=\\displaystyle\\int \\left(\\frac{1}{x+1}+\\frac{2}{x - 2}\\right)dx[\/latex].<\/div>\r\n<p id=\"fs-id1165041765196\">The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of a rational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of the denominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition can be applied to a rational function [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] only if [latex]\\text{deg}\\left(P\\left(x\\right)\\right)&lt;\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex]. In the case when [latex]\\text{deg}\\left(P\\left(x\\right)\\right)\\ge \\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex], we must first perform long division to rewrite the quotient [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] in the form [latex]A\\left(x\\right)+\\frac{R\\left(x\\right)}{Q\\left(x\\right)}[\/latex], where [latex]\\text{deg}\\left(R\\left(x\\right)\\right)&lt;\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex]. We then do a partial fraction decomposition on [latex]\\frac{R\\left(x\\right)}{Q\\left(x\\right)}[\/latex]. The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form [latex]\\displaystyle\\int \\frac{P\\left(x\\right)}{Q\\left(x\\right)}dx[\/latex], where [latex]\\text{deg}\\left(P\\left(x\\right)\\right)\\ge \\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165042228991\" data-type=\"example\">\r\n<div id=\"fs-id1165040798304\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042044846\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Integrating [latex]\\displaystyle\\int \\frac{P\\left(x\\right)}{Q\\left(x\\right)}dx[\/latex], where [latex]\\text{deg}\\left(P\\left(x\\right)\\right)\\ge \\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex]<\/h3>\r\n<div id=\"fs-id1165042044846\" data-type=\"problem\">\r\n<p id=\"fs-id1165041845429\">Evaluate [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+5}{x+1}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1165040794978\" data-type=\"solution\">\r\n<p id=\"fs-id1165041796692\">Since [latex]\\text{deg}\\left({x}^{2}+3x+5\\right)\\ge \\text{deg}\\left(x+1\\right)[\/latex], we perform long division to obtain<\/p>\r\n\r\n<div id=\"fs-id1165040774466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{2}+3x+5}{x+1}=x+2+\\frac{3}{x+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040796470\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042278063\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int \\frac{{x}^{2}+3x+5}{x+1}dx}&amp; ={\\displaystyle\\int \\left(x+2+\\frac{3}{x+1}\\right)dx}\\hfill \\\\ &amp; =\\frac{1}{2}{x}^{2}+2x+3\\text{ln}|x+1|+C.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165041820765\" class=\"media-2\" data-type=\"note\">\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Polynomial Long Division<\/h3>\r\n<ol>\r\n \t<li>Set up the division problem as the numerator divided by the denominator<\/li>\r\n \t<li>Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.<\/li>\r\n \t<li>Multiply the answer by the divisor and write it below the like terms of the dividend.<\/li>\r\n \t<li>Subtract the bottom binomial from the top binomial.<\/li>\r\n \t<li>Bring down the next term of the dividend.<\/li>\r\n \t<li>Repeat steps 2\u20135 until reaching the last term of the dividend.<\/li>\r\n \t<li>If the remainder is non-zero, express as a fraction using the divisor as the denominator.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<p id=\"fs-id1165041821545\">Visit <a href=\"https:\/\/en.wikipedia.org\/wiki\/Polynomial_long_division\" target=\"_blank\" rel=\"noopener\">this website for a review of long division of polynomials<\/a>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165041972615\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042281186\" data-type=\"exercise\">\r\n<div id=\"fs-id1165040743552\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165040743552\" data-type=\"problem\">\r\n<p id=\"fs-id1165042047432\">Evaluate [latex]\\displaystyle\\int \\frac{x - 3}{x+2}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1165042133890\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165041977487\">Use long division to obtain [latex]\\frac{x - 3}{x+2}=1-\\frac{5}{x+2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1165041985161\" data-type=\"solution\">\r\n<p id=\"fs-id1165042050319\">[latex]x - 5\\text{ln}|x+2|+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/STtozLX2gbk?controls=0&amp;start=135&amp;end=182&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.4PartialFractions135to182_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.4 Partial Fractions\" here (opens in new window)<\/a>.\r\n<p id=\"fs-id1165042110224\">To integrate [latex]\\displaystyle\\int \\frac{P\\left(x\\right)}{Q\\left(x\\right)}dx[\/latex], where [latex]\\text{deg}\\left(P\\left(x\\right)\\right)&lt;\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex], we must begin by factoring [latex]Q\\left(x\\right)[\/latex].<\/p>\r\n\r\n<section id=\"fs-id1165042276556\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Nonrepeated Linear Factors<\/h2>\r\n<p id=\"fs-id1165042092059\">If [latex]Q\\left(x\\right)[\/latex] can be factored as [latex]\\left({a}_{1}x+{b}_{1}\\right)\\left({a}_{2}x+{b}_{2}\\right)\\ldots\\left({a}_{n}x+{b}_{n}\\right)[\/latex], where each linear factor is distinct, then it is possible to find constants [latex]{A}_{1},{A}_{2}\\text{,}\\ldots {A}_{n}[\/latex] satisfying<\/p>\r\n\r\n<div id=\"fs-id1165042004560\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{A}_{1}}{{a}_{1}x+{b}_{1}}+\\frac{{A}_{2}}{{a}_{2}x+{b}_{2}}+\\cdots +\\frac{{A}_{n}}{{a}_{n}x+{b}_{n}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042003843\">The proof that such constants exist is beyond the scope of this course.<\/p>\r\n<p id=\"fs-id1165040795049\">In this next example, we see how to use partial fractions to integrate a rational function of this type.<\/p>\r\n\r\n<div id=\"fs-id1165042007126\" data-type=\"example\">\r\n<div id=\"fs-id1165040755188\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042126842\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Partial Fractions with Nonrepeated Linear Factors<\/h3>\r\n<div id=\"fs-id1165042126842\" data-type=\"problem\">\r\n<p id=\"fs-id1165041813414\">Evaluate [latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1165042303798\" data-type=\"solution\">\r\n<p id=\"fs-id1165042018008\">Since [latex]\\text{deg}\\left(3x+2\\right)&lt;\\text{deg}\\left({x}^{3}-{x}^{2}-2x\\right)[\/latex], we begin by factoring the denominator of [latex]\\frac{3x+2}{{x}^{3}-{x}^{2}-2x}[\/latex]. We can see that [latex]{x}^{3}-{x}^{2}-2x=x\\left(x - 2\\right)\\left(x+1\\right)[\/latex]. Thus, there are constants [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] satisfying<\/p>\r\n\r\n<div id=\"fs-id1165040669618\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{3x+2}{x\\left(x - 2\\right)\\left(x+1\\right)}=\\frac{A}{x}+\\frac{B}{x - 2}+\\frac{C}{x+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041915383\">We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165041977500\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{3x+2}{x\\left(x - 2\\right)\\left(x+1\\right)}=\\frac{A\\left(x - 2\\right)\\left(x+1\\right)+Bx\\left(x+1\\right)+Cx\\left(x - 2\\right)}{x\\left(x - 2\\right)\\left(x+1\\right)}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042049257\">Now, we set the numerators equal to each other, obtaining<\/p>\r\n\r\n<div id=\"fs-id1165042002695\" style=\"text-align: center;\" data-type=\"equation\">[latex]3x+2=A\\left(x - 2\\right)\\left(x+1\\right)+Bx\\left(x+1\\right)+Cx\\left(x - 2\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041768242\">There are two different strategies for finding the coefficients [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex]. We refer to these as the <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">method of equating coefficients<\/em><\/span> and the <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">method of strategic substitution<\/em><\/span>.<\/p>\r\n\r\n<div id=\"fs-id1165040740860\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox\">\r\n<h2 data-type=\"title\">Rule: Method of Equating Coefficients<\/h2>\r\n<p id=\"fs-id1165041841985\">Rewrite the previous equation in the form<\/p>\r\n\r\n<div id=\"fs-id1165040797991\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3x+2=\\left(A+B+C\\right){x}^{2}+\\left(\\text{-}A+B - 2C\\right)x+\\left(-2A\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042090588\">Equating coefficients produces the system of equations<\/p>\r\n\r\n<div id=\"fs-id1165041818055\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill A+B+C&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill -A+B - 2C&amp; =\\hfill &amp; 3\\hfill \\\\ \\hfill -2A&amp; =\\hfill &amp; 2.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042094196\">To solve this system, we first observe that [latex]-2A=2\\Rightarrow A=-1[\/latex]. Substituting this value into the first two equations gives us the system<\/p>\r\n\r\n<div id=\"fs-id1165040757789\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill B+C&amp; =\\hfill &amp; 1\\hfill \\\\ \\hfill B - 2C&amp; =\\hfill &amp; 2.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041892413\">Multiplying the second equation by [latex]-1[\/latex] and adding the resulting equation to the first produces<\/p>\r\n\r\n<div id=\"fs-id1165041843249\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]-3C=1[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042232940\">which in turn implies that [latex]C=-\\frac{1}{3}[\/latex]. Substituting this value into the equation [latex]B+C=1[\/latex] yields [latex]B=\\frac{4}{3}[\/latex]. Thus, solving these equations yields [latex]A=-1[\/latex], [latex]B=\\frac{4}{3}[\/latex], and [latex]C=-\\frac{1}{3}[\/latex].<\/p>\r\n<p id=\"fs-id1165040743265\">It is important to note that the system produced by this method is consistent if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165040763120\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox\">\r\n<h2 data-type=\"title\">Rule: Method of Strategic Substitution<\/h2>\r\n<p id=\"fs-id1165042220434\">The method of strategic substitution is based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] that satisfy the equation for <em data-effect=\"italics\">all<\/em> values of [latex]x[\/latex]. That is, this equation must be true for any value of [latex]x[\/latex] we care to substitute into it. Therefore, by choosing values of [latex]x[\/latex] carefully and substituting them into the equation, we may find [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] easily. For example, if we substitute [latex]x=0[\/latex], the equation reduces to [latex]2=A\\left(-2\\right)\\left(1\\right)[\/latex]. Solving for [latex]A[\/latex] yields [latex]A=-1[\/latex]. Next, by substituting [latex]x=2[\/latex], the equation reduces to [latex]8=B\\left(2\\right)\\left(3\\right)[\/latex], or equivalently [latex]B=\\frac{4}{3}[\/latex]. Last, we substitute [latex]x=-1[\/latex] into the equation and obtain [latex]-1=C\\left(-1\\right)\\left(-3\\right)[\/latex]. Solving, we have [latex]C=-\\frac{1}{3}[\/latex].<\/p>\r\n<p id=\"fs-id1165040668264\">It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165040668267\">Now that we have the values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex], we rewrite the original integral:<\/p>\r\n\r\n<div id=\"fs-id1165042110103\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\\displaystyle\\int \\left(\\text{-}\\frac{1}{x}+\\frac{4}{3}\\cdot \\frac{1}{\\left(x - 2\\right)}-\\frac{1}{3}\\cdot \\frac{1}{\\left(x+1\\right)}\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042048571\">Evaluating the integral gives us<\/p>\r\n\r\n<div id=\"fs-id1165041985166\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\\text{-}\\text{ln}|x|+\\frac{4}{3}\\text{ln}|x - 2|-\\frac{1}{3}\\text{ln}|x+1|+C[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165041845305\">In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of the denominator.<\/p>\r\n\r\n<div id=\"fs-id1165042066339\" data-type=\"example\">\r\n<div id=\"fs-id1165040797346\" data-type=\"exercise\">\r\n<div id=\"fs-id1165040797348\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Dividing before Applying Partial Fractions<\/h3>\r\n<div id=\"fs-id1165040797348\" data-type=\"problem\">\r\n<p id=\"fs-id1165042042169\">Evaluate [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1165042236262\" data-type=\"solution\">\r\n<p id=\"fs-id1165042234315\">Since [latex]\\text{degree}\\left({x}^{2}+3x+1\\right)\\ge \\text{degree}\\left({x}^{2}-4\\right)[\/latex], we must perform long division of polynomials. This results in<\/p>\r\n\r\n<div id=\"fs-id1165042277396\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{2}+3x+1}{{x}^{2}-4}=1+\\frac{3x+5}{{x}^{2}-4}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042280282\">Next, we perform partial fraction decomposition on [latex]\\frac{3x+5}{{x}^{2}-4}=\\frac{3x+5}{\\left(x+2\\right)\\left(x - 2\\right)}[\/latex]. We have<\/p>\r\n\r\n<div id=\"fs-id1165042092094\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{3x+5}{\\left(x - 2\\right)\\left(x+2\\right)}=\\frac{A}{x - 2}+\\frac{B}{x+2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041757490\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042134742\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3x+5=A\\left(x+2\\right)+B\\left(x - 2\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040641465\">Solving for [latex]A[\/latex] and [latex]B[\/latex] using either method, we obtain [latex]A=\\frac{11}{4}[\/latex] and [latex]B=\\frac{1}{4}[\/latex].<\/p>\r\n<p id=\"fs-id1165041763604\">Rewriting the original integral, we have<\/p>\r\n\r\n<div id=\"fs-id1165041840380\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=\\displaystyle\\int \\left(1+\\frac{11}{4}\\cdot \\frac{1}{x - 2}+\\frac{1}{4}\\cdot \\frac{1}{x+2}\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042050638\">Evaluating the integral produces<\/p>\r\n\r\n<div id=\"fs-id1165042050641\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=x+\\frac{11}{4}\\text{ln}|x - 2|+\\frac{1}{4}\\text{ln}|x+2|+C[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165042280920\">As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. The trick is to convert the nonrational function to a rational function through a substitution.<\/p>\r\n\r\n<div id=\"fs-id1165042003975\" data-type=\"example\">\r\n<div id=\"fs-id1165042003977\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042003980\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Applying Partial Fractions after a Substitution<\/h3>\r\n<div id=\"fs-id1165042003980\" data-type=\"problem\">\r\n<p id=\"fs-id1165042277642\">Evaluate [latex]\\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1165040641550\" data-type=\"solution\">\r\n<p id=\"fs-id1165040641552\">Let\u2019s begin by letting [latex]u=\\sin{x}[\/latex]. Consequently, [latex]du=\\cos{x}dx[\/latex]. After making these substitutions, we have<\/p>\r\n\r\n<div id=\"fs-id1165042035790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx=\\displaystyle\\int \\frac{du}{{u}^{2}-u}=\\displaystyle\\int \\frac{du}{u\\left(u - 1\\right)}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041831988\">Applying partial fraction decomposition to [latex]\\frac{1}{u}\\left(u - 1\\right)[\/latex] gives [latex]\\frac{1}{u\\left(u - 1\\right)}=-\\frac{1}{u}+\\frac{1}{u - 1}[\/latex].<\/p>\r\n<p id=\"fs-id1165042126853\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042126856\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx&amp; =\\text{-}\\text{ln}|u|+\\text{ln}|u - 1|+C\\hfill \\\\ &amp; =\\text{-}\\text{ln}|\\sin{x}|+\\text{ln}|\\sin{x} - 1|+C.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042048889\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042048892\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042266568\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165042266568\" data-type=\"problem\">\r\n<p id=\"fs-id1165042266570\">Evaluate [latex]\\displaystyle\\int \\frac{x+1}{\\left(x+3\\right)\\left(x - 2\\right)}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1165042277051\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165041836267\">[latex]\\frac{x+1}{\\left(x+3\\right)\\left(x - 2\\right)}=\\frac{A}{x+3}+\\frac{B}{x - 2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1165042305329\" data-type=\"solution\">\r\n<p id=\"fs-id1165042305331\">[latex]\\frac{2}{5}\\text{ln}|x+3|+\\frac{3}{5}\\text{ln}|x - 2|+C[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/STtozLX2gbk?controls=0&amp;start=826&amp;end=953&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.4PartialFractions826to953_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.4 Partial Fractions\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section id=\"fs-id1165040741023\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Repeated Linear Factors<\/h2>\r\n<p id=\"fs-id1165040768410\">For some applications, we need to integrate rational expressions that have denominators with repeated linear factors\u2014that is, rational functions with at least one factor of the form [latex]{\\left(ax+b\\right)}^{n}[\/latex], where [latex]n[\/latex] is a positive integer greater than or equal to [latex]2[\/latex]. If the denominator contains the repeated linear factor [latex]{\\left(ax+b\\right)}^{n}[\/latex], then the decomposition must contain<\/p>\r\n\r\n<div id=\"fs-id1165041924064\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{A}_{1}}{ax+b}+\\frac{{A}_{2}}{{\\left(ax+b\\right)}^{2}}+\\cdots +\\frac{{A}_{n}}{{\\left(ax+b\\right)}^{n}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040744273\">As we see in our next example, the basic technique used for solving for the coefficients is the same, but it requires more algebra to determine the numerators of the partial fractions.<\/p>\r\n\r\n<div id=\"fs-id1165041952248\" data-type=\"example\">\r\n<div id=\"fs-id1165041952250\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042094506\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Partial Fractions with Repeated Linear Factors<\/h3>\r\n<div id=\"fs-id1165042094506\" data-type=\"problem\">\r\n<p id=\"fs-id1165040688164\">Evaluate [latex]\\displaystyle\\int \\frac{x - 2}{{\\left(2x - 1\\right)}^{2}\\left(x - 1\\right)}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1165041999301\" data-type=\"solution\">\r\n<p id=\"fs-id1165041804284\">We have [latex]\\text{degree}\\left(x - 2\\right)&lt;\\text{degree}\\left({\\left(2x - 1\\right)}^{2}\\left(x - 1\\right)\\right)[\/latex], so we can proceed with the decomposition. Since [latex]{\\left(2x - 1\\right)}^{2}[\/latex] is a repeated linear factor, include [latex]\\frac{A}{2x - 1}+\\frac{B}{{\\left(2x - 1\\right)}^{2}}[\/latex] in the decomposition. Thus,<\/p>\r\n\r\n<div id=\"fs-id1165040688174\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{x - 2}{{\\left(2x - 1\\right)}^{2}\\left(x - 1\\right)}=\\frac{A}{2x - 1}+\\frac{B}{{\\left(2x - 1\\right)}^{2}}+\\frac{C}{x - 1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040688636\">After getting a common denominator and equating the numerators, we have<\/p>\r\n\r\n<div id=\"fs-id1165040688639\" style=\"text-align: center;\" data-type=\"equation\">[latex]x - 2=A\\left(2x - 1\\right)\\left(x - 1\\right)+B\\left(x - 1\\right)+C{\\left(2x - 1\\right)}^{2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042007681\">We then use the method of equating coefficients to find the values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165040758310\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x - 2=\\left(2A+4C\\right){x}^{2}+\\left(-3A+B - 4C\\right)x+\\left(A-B+C\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042088386\">Equating coefficients yields [latex]2A+4C=0[\/latex], [latex]-3A+B - 4C=1[\/latex], and [latex]A-B+C=-2[\/latex]. Solving this system yields [latex]A=2[\/latex], [latex]B=3[\/latex], and [latex]C=-1[\/latex].<\/p>\r\n<p id=\"fs-id1165042087142\">Alternatively, we can use the method of strategic substitution. In this case, substituting [latex]x=1[\/latex] and [latex]x=\\frac{1}{2}[\/latex] into the previous equation easily produces the values [latex]B=3[\/latex] and [latex]C=-1[\/latex]. At this point, it may seem that we have run out of good choices for [latex]x[\/latex], however, since we already have values for [latex]B[\/latex] and [latex]C[\/latex], we can substitute in these values and choose any value for [latex]x[\/latex] not previously used. The value [latex]x=0[\/latex] is a good option. In this case, we obtain the equation [latex]-2=A\\left(-1\\right)\\left(-1\\right)+3\\left(-1\\right)+\\left(-1\\right){\\left(-1\\right)}^{2}[\/latex] or, equivalently, [latex]A=2[\/latex].<\/p>\r\n<p id=\"fs-id1165042031880\">Now that we have the values for [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex], we rewrite the original integral and evaluate it:<\/p>\r\n\r\n<div id=\"fs-id1165041837274\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int \\frac{x - 2}{{\\left(2x - 1\\right)}^{2}\\left(x - 1\\right)}dx}&amp; ={\\displaystyle\\int \\left(\\frac{2}{2x - 1}+\\frac{3}{{\\left(2x - 1\\right)}^{2}}-\\frac{1}{x - 1}\\right)dx}\\hfill \\\\ &amp; =\\text{ln}|2x - 1|-\\frac{3}{2\\left(2x - 1\\right)}-\\text{ln}|x - 1|+C.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042088541\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165042088545\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042088547\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<div id=\"fs-id1165042088547\" data-type=\"problem\">\r\n<p id=\"fs-id1165042281590\">Set up the partial fraction decomposition for [latex]\\displaystyle\\int \\frac{x+2}{{\\left(x+3\\right)}^{3}{\\left(x - 4\\right)}^{2}}dx[\/latex]. (Do not solve for the coefficients or complete the integration.)<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1165042004334\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165041977514\">Use the problem-solving method of the above example for guidance.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1165042194352\" data-type=\"solution\">\r\n<p id=\"fs-id1165040665662\">[latex]\\frac{x+2}{{\\left(x+3\\right)}^{3}{\\left(x - 4\\right)}^{2}}=\\frac{A}{x+3}+\\frac{B}{{\\left(x+3\\right)}^{2}}+\\frac{C}{{\\left(x+3\\right)}^{3}}+\\frac{D}{\\left(x - 4\\right)}+\\frac{E}{{\\left(x - 4\\right)}^{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/STtozLX2gbk?controls=0&amp;start=1372&amp;end=1422&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.4PartialFractions1372to1422_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.4 Partial Fractions\" here (opens in new window)<\/a>.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Integrate a rational function using the method of partial fractions<\/li>\n<li>Recognize simple linear factors in a rational function<\/li>\n<li>Recognize repeated linear factors in a rational function<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165042047881\">We have seen some techniques that allow us to integrate specific rational functions. For example, we know that<\/p>\n<div id=\"fs-id1165041842105\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{du}{u}=\\text{ln}|u|+C\\text{ and }\\displaystyle\\int \\frac{du}{{u}^{2}+{a}^{2}}=\\frac{1}{a}{\\tan}^{-1}\\left(\\frac{u}{a}\\right)+C\\text{.}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041797885\">However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately obvious how to go about evaluating [latex]\\displaystyle\\int \\frac{3x}{{x}^{2}-x - 2}dx[\/latex]. However, we know from material previously developed that<\/p>\n<div id=\"fs-id1165042089867\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\left(\\frac{1}{x+1}+\\frac{2}{x - 2}\\right)dx=\\text{ln}|x+1|+2\\text{ln}|x - 2|+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041972469\">In fact, by getting a common denominator, we see that<\/p>\n<div id=\"fs-id1165040736457\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{x+1}+\\frac{2}{x - 2}=\\frac{3x}{{x}^{2}-x - 2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041913720\">Consequently,<\/p>\n<div id=\"fs-id1165042275619\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{3x}{{x}^{2}-x - 2}dx=\\displaystyle\\int \\left(\\frac{1}{x+1}+\\frac{2}{x - 2}\\right)dx[\/latex].<\/div>\n<p id=\"fs-id1165041765196\">The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of a rational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of the denominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition can be applied to a rational function [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] only if [latex]\\text{deg}\\left(P\\left(x\\right)\\right)<\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex]. In the case when [latex]\\text{deg}\\left(P\\left(x\\right)\\right)\\ge \\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex], we must first perform long division to rewrite the quotient [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex] in the form [latex]A\\left(x\\right)+\\frac{R\\left(x\\right)}{Q\\left(x\\right)}[\/latex], where [latex]\\text{deg}\\left(R\\left(x\\right)\\right)<\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex]. We then do a partial fraction decomposition on [latex]\\frac{R\\left(x\\right)}{Q\\left(x\\right)}[\/latex]. The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form [latex]\\displaystyle\\int \\frac{P\\left(x\\right)}{Q\\left(x\\right)}dx[\/latex], where [latex]\\text{deg}\\left(P\\left(x\\right)\\right)\\ge \\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex].<\/p>\n<div id=\"fs-id1165042228991\" data-type=\"example\">\n<div id=\"fs-id1165040798304\" data-type=\"exercise\">\n<div id=\"fs-id1165042044846\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Integrating [latex]\\displaystyle\\int \\frac{P\\left(x\\right)}{Q\\left(x\\right)}dx[\/latex], where [latex]\\text{deg}\\left(P\\left(x\\right)\\right)\\ge \\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex]<\/h3>\n<div id=\"fs-id1165042044846\" data-type=\"problem\">\n<p id=\"fs-id1165041845429\">Evaluate [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+5}{x+1}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040794978\" data-type=\"solution\">\n<p id=\"fs-id1165041796692\">Since [latex]\\text{deg}\\left({x}^{2}+3x+5\\right)\\ge \\text{deg}\\left(x+1\\right)[\/latex], we perform long division to obtain<\/p>\n<div id=\"fs-id1165040774466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{2}+3x+5}{x+1}=x+2+\\frac{3}{x+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040796470\">Thus,<\/p>\n<div id=\"fs-id1165042278063\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int \\frac{{x}^{2}+3x+5}{x+1}dx}& ={\\displaystyle\\int \\left(x+2+\\frac{3}{x+1}\\right)dx}\\hfill \\\\ & =\\frac{1}{2}{x}^{2}+2x+3\\text{ln}|x+1|+C.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165041820765\" class=\"media-2\" data-type=\"note\">\n<div class=\"textbox examples\">\n<h3>Recall: Polynomial Long Division<\/h3>\n<ol>\n<li>Set up the division problem as the numerator divided by the denominator<\/li>\n<li>Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.<\/li>\n<li>Multiply the answer by the divisor and write it below the like terms of the dividend.<\/li>\n<li>Subtract the bottom binomial from the top binomial.<\/li>\n<li>Bring down the next term of the dividend.<\/li>\n<li>Repeat steps 2\u20135 until reaching the last term of the dividend.<\/li>\n<li>If the remainder is non-zero, express as a fraction using the divisor as the denominator.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p id=\"fs-id1165041821545\">Visit <a href=\"https:\/\/en.wikipedia.org\/wiki\/Polynomial_long_division\" target=\"_blank\" rel=\"noopener\">this website for a review of long division of polynomials<\/a>.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165041972615\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042281186\" data-type=\"exercise\">\n<div id=\"fs-id1165040743552\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165040743552\" data-type=\"problem\">\n<p id=\"fs-id1165042047432\">Evaluate [latex]\\displaystyle\\int \\frac{x - 3}{x+2}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042133890\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165041977487\">Use long division to obtain [latex]\\frac{x - 3}{x+2}=1-\\frac{5}{x+2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041985161\" data-type=\"solution\">\n<p id=\"fs-id1165042050319\">[latex]x - 5\\text{ln}|x+2|+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/STtozLX2gbk?controls=0&amp;start=135&amp;end=182&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.4PartialFractions135to182_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.4 Partial Fractions&#8221; here (opens in new window)<\/a>.<\/p>\n<p id=\"fs-id1165042110224\">To integrate [latex]\\displaystyle\\int \\frac{P\\left(x\\right)}{Q\\left(x\\right)}dx[\/latex], where [latex]\\text{deg}\\left(P\\left(x\\right)\\right)<\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex], we must begin by factoring [latex]Q\\left(x\\right)[\/latex].<\/p>\n<section id=\"fs-id1165042276556\" data-depth=\"1\">\n<h2 data-type=\"title\">Nonrepeated Linear Factors<\/h2>\n<p id=\"fs-id1165042092059\">If [latex]Q\\left(x\\right)[\/latex] can be factored as [latex]\\left({a}_{1}x+{b}_{1}\\right)\\left({a}_{2}x+{b}_{2}\\right)\\ldots\\left({a}_{n}x+{b}_{n}\\right)[\/latex], where each linear factor is distinct, then it is possible to find constants [latex]{A}_{1},{A}_{2}\\text{,}\\ldots {A}_{n}[\/latex] satisfying<\/p>\n<div id=\"fs-id1165042004560\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{A}_{1}}{{a}_{1}x+{b}_{1}}+\\frac{{A}_{2}}{{a}_{2}x+{b}_{2}}+\\cdots +\\frac{{A}_{n}}{{a}_{n}x+{b}_{n}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042003843\">The proof that such constants exist is beyond the scope of this course.<\/p>\n<p id=\"fs-id1165040795049\">In this next example, we see how to use partial fractions to integrate a rational function of this type.<\/p>\n<div id=\"fs-id1165042007126\" data-type=\"example\">\n<div id=\"fs-id1165040755188\" data-type=\"exercise\">\n<div id=\"fs-id1165042126842\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Partial Fractions with Nonrepeated Linear Factors<\/h3>\n<div id=\"fs-id1165042126842\" data-type=\"problem\">\n<p id=\"fs-id1165041813414\">Evaluate [latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042303798\" data-type=\"solution\">\n<p id=\"fs-id1165042018008\">Since [latex]\\text{deg}\\left(3x+2\\right)<\\text{deg}\\left({x}^{3}-{x}^{2}-2x\\right)[\/latex], we begin by factoring the denominator of [latex]\\frac{3x+2}{{x}^{3}-{x}^{2}-2x}[\/latex]. We can see that [latex]{x}^{3}-{x}^{2}-2x=x\\left(x - 2\\right)\\left(x+1\\right)[\/latex]. Thus, there are constants [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] satisfying<\/p>\n<div id=\"fs-id1165040669618\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{3x+2}{x\\left(x - 2\\right)\\left(x+1\\right)}=\\frac{A}{x}+\\frac{B}{x - 2}+\\frac{C}{x+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041915383\">We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,<\/p>\n<div id=\"fs-id1165041977500\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{3x+2}{x\\left(x - 2\\right)\\left(x+1\\right)}=\\frac{A\\left(x - 2\\right)\\left(x+1\\right)+Bx\\left(x+1\\right)+Cx\\left(x - 2\\right)}{x\\left(x - 2\\right)\\left(x+1\\right)}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042049257\">Now, we set the numerators equal to each other, obtaining<\/p>\n<div id=\"fs-id1165042002695\" style=\"text-align: center;\" data-type=\"equation\">[latex]3x+2=A\\left(x - 2\\right)\\left(x+1\\right)+Bx\\left(x+1\\right)+Cx\\left(x - 2\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041768242\">There are two different strategies for finding the coefficients [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex]. We refer to these as the <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">method of equating coefficients<\/em><\/span> and the <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">method of strategic substitution<\/em><\/span>.<\/p>\n<div id=\"fs-id1165040740860\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox\">\n<h2 data-type=\"title\">Rule: Method of Equating Coefficients<\/h2>\n<p id=\"fs-id1165041841985\">Rewrite the previous equation in the form<\/p>\n<div id=\"fs-id1165040797991\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3x+2=\\left(A+B+C\\right){x}^{2}+\\left(\\text{-}A+B - 2C\\right)x+\\left(-2A\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042090588\">Equating coefficients produces the system of equations<\/p>\n<div id=\"fs-id1165041818055\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill A+B+C& =\\hfill & 0\\hfill \\\\ \\hfill -A+B - 2C& =\\hfill & 3\\hfill \\\\ \\hfill -2A& =\\hfill & 2.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042094196\">To solve this system, we first observe that [latex]-2A=2\\Rightarrow A=-1[\/latex]. Substituting this value into the first two equations gives us the system<\/p>\n<div id=\"fs-id1165040757789\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill B+C& =\\hfill & 1\\hfill \\\\ \\hfill B - 2C& =\\hfill & 2.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041892413\">Multiplying the second equation by [latex]-1[\/latex] and adding the resulting equation to the first produces<\/p>\n<div id=\"fs-id1165041843249\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]-3C=1[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042232940\">which in turn implies that [latex]C=-\\frac{1}{3}[\/latex]. Substituting this value into the equation [latex]B+C=1[\/latex] yields [latex]B=\\frac{4}{3}[\/latex]. Thus, solving these equations yields [latex]A=-1[\/latex], [latex]B=\\frac{4}{3}[\/latex], and [latex]C=-\\frac{1}{3}[\/latex].<\/p>\n<p id=\"fs-id1165040743265\">It is important to note that the system produced by this method is consistent if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165040763120\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox\">\n<h2 data-type=\"title\">Rule: Method of Strategic Substitution<\/h2>\n<p id=\"fs-id1165042220434\">The method of strategic substitution is based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] that satisfy the equation for <em data-effect=\"italics\">all<\/em> values of [latex]x[\/latex]. That is, this equation must be true for any value of [latex]x[\/latex] we care to substitute into it. Therefore, by choosing values of [latex]x[\/latex] carefully and substituting them into the equation, we may find [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] easily. For example, if we substitute [latex]x=0[\/latex], the equation reduces to [latex]2=A\\left(-2\\right)\\left(1\\right)[\/latex]. Solving for [latex]A[\/latex] yields [latex]A=-1[\/latex]. Next, by substituting [latex]x=2[\/latex], the equation reduces to [latex]8=B\\left(2\\right)\\left(3\\right)[\/latex], or equivalently [latex]B=\\frac{4}{3}[\/latex]. Last, we substitute [latex]x=-1[\/latex] into the equation and obtain [latex]-1=C\\left(-1\\right)\\left(-3\\right)[\/latex]. Solving, we have [latex]C=-\\frac{1}{3}[\/latex].<\/p>\n<p id=\"fs-id1165040668264\">It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165040668267\">Now that we have the values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex], we rewrite the original integral:<\/p>\n<div id=\"fs-id1165042110103\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\\displaystyle\\int \\left(\\text{-}\\frac{1}{x}+\\frac{4}{3}\\cdot \\frac{1}{\\left(x - 2\\right)}-\\frac{1}{3}\\cdot \\frac{1}{\\left(x+1\\right)}\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042048571\">Evaluating the integral gives us<\/p>\n<div id=\"fs-id1165041985166\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\\text{-}\\text{ln}|x|+\\frac{4}{3}\\text{ln}|x - 2|-\\frac{1}{3}\\text{ln}|x+1|+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165041845305\">In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of the denominator.<\/p>\n<div id=\"fs-id1165042066339\" data-type=\"example\">\n<div id=\"fs-id1165040797346\" data-type=\"exercise\">\n<div id=\"fs-id1165040797348\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Dividing before Applying Partial Fractions<\/h3>\n<div id=\"fs-id1165040797348\" data-type=\"problem\">\n<p id=\"fs-id1165042042169\">Evaluate [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042236262\" data-type=\"solution\">\n<p id=\"fs-id1165042234315\">Since [latex]\\text{degree}\\left({x}^{2}+3x+1\\right)\\ge \\text{degree}\\left({x}^{2}-4\\right)[\/latex], we must perform long division of polynomials. This results in<\/p>\n<div id=\"fs-id1165042277396\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{2}+3x+1}{{x}^{2}-4}=1+\\frac{3x+5}{{x}^{2}-4}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042280282\">Next, we perform partial fraction decomposition on [latex]\\frac{3x+5}{{x}^{2}-4}=\\frac{3x+5}{\\left(x+2\\right)\\left(x - 2\\right)}[\/latex]. We have<\/p>\n<div id=\"fs-id1165042092094\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{3x+5}{\\left(x - 2\\right)\\left(x+2\\right)}=\\frac{A}{x - 2}+\\frac{B}{x+2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041757490\">Thus,<\/p>\n<div id=\"fs-id1165042134742\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3x+5=A\\left(x+2\\right)+B\\left(x - 2\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040641465\">Solving for [latex]A[\/latex] and [latex]B[\/latex] using either method, we obtain [latex]A=\\frac{11}{4}[\/latex] and [latex]B=\\frac{1}{4}[\/latex].<\/p>\n<p id=\"fs-id1165041763604\">Rewriting the original integral, we have<\/p>\n<div id=\"fs-id1165041840380\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=\\displaystyle\\int \\left(1+\\frac{11}{4}\\cdot \\frac{1}{x - 2}+\\frac{1}{4}\\cdot \\frac{1}{x+2}\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042050638\">Evaluating the integral produces<\/p>\n<div id=\"fs-id1165042050641\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=x+\\frac{11}{4}\\text{ln}|x - 2|+\\frac{1}{4}\\text{ln}|x+2|+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165042280920\">As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. The trick is to convert the nonrational function to a rational function through a substitution.<\/p>\n<div id=\"fs-id1165042003975\" data-type=\"example\">\n<div id=\"fs-id1165042003977\" data-type=\"exercise\">\n<div id=\"fs-id1165042003980\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Applying Partial Fractions after a Substitution<\/h3>\n<div id=\"fs-id1165042003980\" data-type=\"problem\">\n<p id=\"fs-id1165042277642\">Evaluate [latex]\\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Show Solution<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040641550\" data-type=\"solution\">\n<p id=\"fs-id1165040641552\">Let\u2019s begin by letting [latex]u=\\sin{x}[\/latex]. Consequently, [latex]du=\\cos{x}dx[\/latex]. After making these substitutions, we have<\/p>\n<div id=\"fs-id1165042035790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx=\\displaystyle\\int \\frac{du}{{u}^{2}-u}=\\displaystyle\\int \\frac{du}{u\\left(u - 1\\right)}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041831988\">Applying partial fraction decomposition to [latex]\\frac{1}{u}\\left(u - 1\\right)[\/latex] gives [latex]\\frac{1}{u\\left(u - 1\\right)}=-\\frac{1}{u}+\\frac{1}{u - 1}[\/latex].<\/p>\n<p id=\"fs-id1165042126853\">Thus,<\/p>\n<div id=\"fs-id1165042126856\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx& =\\text{-}\\text{ln}|u|+\\text{ln}|u - 1|+C\\hfill \\\\ & =\\text{-}\\text{ln}|\\sin{x}|+\\text{ln}|\\sin{x} - 1|+C.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042048889\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042048892\" data-type=\"exercise\">\n<div id=\"fs-id1165042266568\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165042266568\" data-type=\"problem\">\n<p id=\"fs-id1165042266570\">Evaluate [latex]\\displaystyle\\int \\frac{x+1}{\\left(x+3\\right)\\left(x - 2\\right)}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Hint<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042277051\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165041836267\">[latex]\\frac{x+1}{\\left(x+3\\right)\\left(x - 2\\right)}=\\frac{A}{x+3}+\\frac{B}{x - 2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042305329\" data-type=\"solution\">\n<p id=\"fs-id1165042305331\">[latex]\\frac{2}{5}\\text{ln}|x+3|+\\frac{3}{5}\\text{ln}|x - 2|+C[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/STtozLX2gbk?controls=0&amp;start=826&amp;end=953&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.4PartialFractions826to953_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.4 Partial Fractions&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section id=\"fs-id1165040741023\" data-depth=\"1\">\n<h2 data-type=\"title\">Repeated Linear Factors<\/h2>\n<p id=\"fs-id1165040768410\">For some applications, we need to integrate rational expressions that have denominators with repeated linear factors\u2014that is, rational functions with at least one factor of the form [latex]{\\left(ax+b\\right)}^{n}[\/latex], where [latex]n[\/latex] is a positive integer greater than or equal to [latex]2[\/latex]. If the denominator contains the repeated linear factor [latex]{\\left(ax+b\\right)}^{n}[\/latex], then the decomposition must contain<\/p>\n<div id=\"fs-id1165041924064\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{A}_{1}}{ax+b}+\\frac{{A}_{2}}{{\\left(ax+b\\right)}^{2}}+\\cdots +\\frac{{A}_{n}}{{\\left(ax+b\\right)}^{n}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040744273\">As we see in our next example, the basic technique used for solving for the coefficients is the same, but it requires more algebra to determine the numerators of the partial fractions.<\/p>\n<div id=\"fs-id1165041952248\" data-type=\"example\">\n<div id=\"fs-id1165041952250\" data-type=\"exercise\">\n<div id=\"fs-id1165042094506\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Partial Fractions with Repeated Linear Factors<\/h3>\n<div id=\"fs-id1165042094506\" data-type=\"problem\">\n<p id=\"fs-id1165040688164\">Evaluate [latex]\\displaystyle\\int \\frac{x - 2}{{\\left(2x - 1\\right)}^{2}\\left(x - 1\\right)}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041999301\" data-type=\"solution\">\n<p id=\"fs-id1165041804284\">We have [latex]\\text{degree}\\left(x - 2\\right)<\\text{degree}\\left({\\left(2x - 1\\right)}^{2}\\left(x - 1\\right)\\right)[\/latex], so we can proceed with the decomposition. Since [latex]{\\left(2x - 1\\right)}^{2}[\/latex] is a repeated linear factor, include [latex]\\frac{A}{2x - 1}+\\frac{B}{{\\left(2x - 1\\right)}^{2}}[\/latex] in the decomposition. Thus,<\/p>\n<div id=\"fs-id1165040688174\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{x - 2}{{\\left(2x - 1\\right)}^{2}\\left(x - 1\\right)}=\\frac{A}{2x - 1}+\\frac{B}{{\\left(2x - 1\\right)}^{2}}+\\frac{C}{x - 1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040688636\">After getting a common denominator and equating the numerators, we have<\/p>\n<div id=\"fs-id1165040688639\" style=\"text-align: center;\" data-type=\"equation\">[latex]x - 2=A\\left(2x - 1\\right)\\left(x - 1\\right)+B\\left(x - 1\\right)+C{\\left(2x - 1\\right)}^{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042007681\">We then use the method of equating coefficients to find the values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex].<\/p>\n<div id=\"fs-id1165040758310\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]x - 2=\\left(2A+4C\\right){x}^{2}+\\left(-3A+B - 4C\\right)x+\\left(A-B+C\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042088386\">Equating coefficients yields [latex]2A+4C=0[\/latex], [latex]-3A+B - 4C=1[\/latex], and [latex]A-B+C=-2[\/latex]. Solving this system yields [latex]A=2[\/latex], [latex]B=3[\/latex], and [latex]C=-1[\/latex].<\/p>\n<p id=\"fs-id1165042087142\">Alternatively, we can use the method of strategic substitution. In this case, substituting [latex]x=1[\/latex] and [latex]x=\\frac{1}{2}[\/latex] into the previous equation easily produces the values [latex]B=3[\/latex] and [latex]C=-1[\/latex]. At this point, it may seem that we have run out of good choices for [latex]x[\/latex], however, since we already have values for [latex]B[\/latex] and [latex]C[\/latex], we can substitute in these values and choose any value for [latex]x[\/latex] not previously used. The value [latex]x=0[\/latex] is a good option. In this case, we obtain the equation [latex]-2=A\\left(-1\\right)\\left(-1\\right)+3\\left(-1\\right)+\\left(-1\\right){\\left(-1\\right)}^{2}[\/latex] or, equivalently, [latex]A=2[\/latex].<\/p>\n<p id=\"fs-id1165042031880\">Now that we have the values for [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex], we rewrite the original integral and evaluate it:<\/p>\n<div id=\"fs-id1165041837274\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int \\frac{x - 2}{{\\left(2x - 1\\right)}^{2}\\left(x - 1\\right)}dx}& ={\\displaystyle\\int \\left(\\frac{2}{2x - 1}+\\frac{3}{{\\left(2x - 1\\right)}^{2}}-\\frac{1}{x - 1}\\right)dx}\\hfill \\\\ & =\\text{ln}|2x - 1|-\\frac{3}{2\\left(2x - 1\\right)}-\\text{ln}|x - 1|+C.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042088541\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165042088545\" data-type=\"exercise\">\n<div id=\"fs-id1165042088547\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<div id=\"fs-id1165042088547\" data-type=\"problem\">\n<p id=\"fs-id1165042281590\">Set up the partial fraction decomposition for [latex]\\displaystyle\\int \\frac{x+2}{{\\left(x+3\\right)}^{3}{\\left(x - 4\\right)}^{2}}dx[\/latex]. (Do not solve for the coefficients or complete the integration.)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Hint<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042004334\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165041977514\">Use the problem-solving method of the above example for guidance.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Show Solution<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042194352\" data-type=\"solution\">\n<p id=\"fs-id1165040665662\">[latex]\\frac{x+2}{{\\left(x+3\\right)}^{3}{\\left(x - 4\\right)}^{2}}=\\frac{A}{x+3}+\\frac{B}{{\\left(x+3\\right)}^{2}}+\\frac{C}{{\\left(x+3\\right)}^{3}}+\\frac{D}{\\left(x - 4\\right)}+\\frac{E}{{\\left(x - 4\\right)}^{2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/STtozLX2gbk?controls=0&amp;start=1372&amp;end=1422&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.4PartialFractions1372to1422_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.4 Partial Fractions&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1586\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.4 Partial Fractions. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.4 Partial Fractions\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1586","chapter","type-chapter","status-publish","hentry"],"part":158,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1586","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1586\/revisions"}],"predecessor-version":[{"id":2031,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1586\/revisions\/2031"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/158"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1586\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1586"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1586"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1586"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1586"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}