{"id":1587,"date":"2021-07-22T16:30:07","date_gmt":"2021-07-22T16:30:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1587"},"modified":"2022-03-19T04:09:37","modified_gmt":"2022-03-19T04:09:37","slug":"partial-fraction-decomposition","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/partial-fraction-decomposition\/","title":{"raw":"Partial Fraction Decomposition","rendered":"Partial Fraction Decomposition"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize quadratic factors in a rational function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1165042042804\" data-depth=\"1\">\r\n<h2>The General Method<\/h2>\r\n<p id=\"fs-id1165042042809\">Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let\u2019s outline the basic method in the following problem-solving strategy.<\/p>\r\n\r\n<div id=\"fs-id1165040639180\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Problem-Solving Strategy: Partial Fraction Decomposition<\/h3>\r\n<p id=\"fs-id1165040639187\">To decompose the rational function [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex], use the following steps:<\/p>\r\n\r\n<ol id=\"fs-id1165042108900\" type=\"1\">\r\n \t<li>Make sure that [latex]\\text{degree}\\left(P\\left(x\\right)\\right)&lt;\\text{degree}\\left(Q\\left(x\\right)\\right)[\/latex]. If not, perform long division of polynomials.<\/li>\r\n \t<li>Factor [latex]Q\\left(x\\right)[\/latex] into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.<\/li>\r\n \t<li>Assuming that [latex]\\text{deg}\\left(P\\left(x\\right)\\right)&lt;\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex], the factors of [latex]Q\\left(x\\right)[\/latex] determine the form of the decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex].\r\n<ol id=\"fs-id1165040671939\" type=\"a\">\r\n \t<li>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},...{A}_{n}[\/latex] satisfying<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165041899414\" 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;<\/li>\r\n \t<li>If [latex]Q\\left(x\\right)[\/latex] contains the repeated linear factor [latex]{\\left(ax+b\\right)}^{n}[\/latex], then the decomposition must contain<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165040643994\" 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;<\/li>\r\n \t<li>For each irreducible quadratic factor [latex]a{x}^{2}+bx+c[\/latex] that [latex]Q\\left(x\\right)[\/latex] contains, the decomposition must include<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165042277015\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{Ax+B}{a{x}^{2}+bx+c}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>For each repeated irreducible quadratic factor [latex]{\\left(a{x}^{2}+bx+c\\right)}^{n}[\/latex], the decomposition must include<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165042230430\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{A}_{1}x+{B}_{1}}{a{x}^{2}+bx+c}+\\frac{{A}_{2}x+{B}_{2}}{{\\left(a{x}^{2}+bx+c\\right)}^{2}}+\\cdots +\\frac{{A}_{n}x+{B}_{n}}{{\\left(a{x}^{2}+bx+c\\right)}^{n}}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>After the appropriate decomposition is determined, solve for the constants.<\/li>\r\n \t<li>Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165040671182\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Simple Quadratic Factors<\/h2>\r\n<p id=\"fs-id1165040671188\">Now let\u2019s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic [latex]a{x}^{2}+bx+c[\/latex] is irreducible if [latex]a{x}^{2}+bx+c=0[\/latex] has no real zeros\u2014that is, if [latex]{b}^{2}-4ac&lt;0[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165042096550\" data-type=\"example\">\r\n<div id=\"fs-id1165042096552\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042096554\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Rational Expressions with an Irreducible Quadratic Factor<\/h3>\r\n<div id=\"fs-id1165042096554\" data-type=\"problem\">\r\n<p id=\"fs-id1165042056695\">Evaluate [latex]\\displaystyle\\int \\frac{2x - 3}{{x}^{3}+x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1165040713993\" data-type=\"solution\">\r\n<p id=\"fs-id1165041966107\">Since [latex]\\text{deg}\\left(2x - 3\\right)&lt;\\text{deg}\\left({x}^{3}+x\\right)[\/latex], factor the denominator and proceed with partial fraction decomposition. Since [latex]{x}^{3}+x=x\\left({x}^{2}+1\\right)[\/latex] contains the irreducible quadratic factor [latex]{x}^{2}+1[\/latex], include [latex]\\frac{Ax+B}{{x}^{2}+1}[\/latex] as part of the decomposition, along with [latex]\\frac{C}{x}[\/latex] for the linear term [latex]x[\/latex]. Thus, the decomposition has the form<\/p>\r\n\r\n<div id=\"fs-id1165042233088\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{2x - 3}{x\\left({x}^{2}+1\\right)}=\\frac{Ax+B}{{x}^{2}+1}+\\frac{C}{x}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040745299\">After getting a common denominator and equating the numerators, we obtain the equation<\/p>\r\n\r\n<div id=\"fs-id1165040745302\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2x - 3=\\left(Ax+B\\right)x+C\\left({x}^{2}+1\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041952135\">Solving for [latex]A,B[\/latex], and [latex]C[\/latex], we get [latex]A=3[\/latex], [latex]B=2[\/latex], and [latex]C=-3[\/latex].<\/p>\r\n<p id=\"fs-id1165042273867\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1165042273871\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{2x - 3}{{x}^{3}+x}=\\frac{3x+2}{{x}^{2}+1}-\\frac{3}{x}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040757650\">Substituting back into the integral, we obtain<\/p>\r\n\r\n<div id=\"fs-id1165040757653\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{2x - 3}{{x}^{3}+x}dx}&amp; ={\\displaystyle\\int \\left(\\frac{3x+2}{{x}^{2}+1}-\\frac{3}{x}\\right)dx}\\hfill &amp; &amp; &amp; \\\\ &amp; =3{\\displaystyle\\int \\frac{x}{{x}^{2}+1}dx+2\\displaystyle\\int \\frac{1}{{x}^{2}+1}dx - 3\\displaystyle\\int \\frac{1}{x}dx}\\hfill &amp; &amp; &amp; \\text{Split up the integral.}\\hfill \\\\ &amp; =\\frac{3}{2}\\text{ln}|{x}^{2}+1|+2{\\tan}^{-1}x - 3\\text{ln}|x|+C.\\hfill &amp; &amp; &amp; \\text{Evaluate each integral.}\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040775841\"><em data-effect=\"italics\">Note<\/em>: We may rewrite [latex]\\text{ln}|{x}^{2}+1|=\\text{ln}\\left({x}^{2}+1\\right)[\/latex], if we wish to do so, since [latex]{x}^{2}+1&gt;0[\/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<\/div>\r\n<div id=\"fs-id1165042107572\" data-type=\"example\">\r\n<div id=\"fs-id1165042107574\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042107576\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Partial Fractions with an Irreducible Quadratic Factor<\/h3>\r\n<div id=\"fs-id1165042107576\" data-type=\"problem\">\r\n<p id=\"fs-id1165042107582\">Evaluate [latex]\\displaystyle\\int \\frac{dx}{{x}^{3}-8}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1165041952824\" data-type=\"solution\">\r\n<p id=\"fs-id1165041952826\">We can start by factoring [latex]{x}^{3}-8=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]. We see that the quadratic factor [latex]{x}^{2}+2x+4[\/latex] is irreducible since [latex]{2}^{2}-4\\left(1\\right)\\left(4\\right)=-12&lt;0[\/latex]. Using the decomposition described in the problem-solving strategy, we get<\/p>\r\n\r\n<div id=\"fs-id1165041874053\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)}=\\frac{A}{x - 2}+\\frac{Bx+C}{{x}^{2}+2x+4}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042238062\">After obtaining a common denominator and equating the numerators, this becomes<\/p>\r\n\r\n<div id=\"fs-id1165042238065\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1=A\\left({x}^{2}+2x+4\\right)+\\left(Bx+C\\right)\\left(x - 2\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041830086\">Applying either method, we get [latex]A=\\frac{1}{12},B=-\\frac{1}{12},\\text{and}C=-\\frac{1}{3}[\/latex].<\/p>\r\n<p id=\"fs-id1165042243673\">Rewriting [latex]\\displaystyle\\int \\frac{dx}{{x}^{3}-8}[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1165042047720\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{dx}{{x}^{3}-8}=\\frac{1}{12}\\displaystyle\\int \\frac{1}{x - 2}dx-\\frac{1}{12}\\displaystyle\\int \\frac{x+4}{{x}^{2}+2x+4}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165041836432\">We can see that<\/p>\r\n<p id=\"fs-id1165041836436\">[latex]\\displaystyle\\int \\frac{1}{x - 2}dx=\\text{ln}|x - 2|+C[\/latex], but [latex]\\displaystyle\\int \\frac{x+4}{{x}^{2}+2x+4}dx[\/latex] requires a bit more effort. Let\u2019s begin by completing the square on [latex]{x}^{2}+2x+4[\/latex] to obtain<\/p>\r\n\r\n<div id=\"fs-id1165040772336\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{x}^{2}+2x+4={\\left(x+1\\right)}^{2}+3[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040638619\">By letting [latex]u=x+1[\/latex] and consequently [latex]du=dx[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1165042092238\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{x+4}{{x}^{2}+2x+4}dx}&amp; ={\\displaystyle\\int \\frac{x+4}{{\\left(x+1\\right)}^{2}+3}dx}\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Complete the square on the}\\hfill \\\\ \\text{denominator.}\\hfill \\end{array}\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{u+3}{{u}^{2}+3}du}\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Substitute}u=x+1,x=u - 1,\\hfill \\\\ \\text{and}du=dx.\\hfill \\end{array}\\hfill \\\\ &amp; ={\\displaystyle\\int \\frac{u}{{u}^{2}+3}du+\\displaystyle\\int \\frac{3}{{u}^{2}+3}du}\\hfill &amp; &amp; &amp; \\text{Split the numerator apart.}\\hfill \\\\ &amp; =\\frac{1}{2}\\text{ln}|{u}^{2}+3|+\\frac{3}{\\sqrt{3}}{\\tan}^{-1}\\frac{u}{\\sqrt{3}}+C\\hfill &amp; &amp; &amp; \\text{Evaluate each integral.}\\hfill \\\\ &amp; =\\frac{1}{2}\\text{ln}|{x}^{2}+2x+4|+\\sqrt{3}{\\tan}^{-1}\\left(\\frac{x+1}{\\sqrt{3}}\\right)+C.\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Rewrite in terms of}x\\text{and}\\hfill \\\\ \\text{simplify.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040772126\">Substituting back into the original integral and simplifying gives<\/p>\r\n\r\n<div id=\"fs-id1165040772130\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int}\\frac{dx}{{x}^{3}-8}=\\frac{1}{12}\\text{ln}|x - 2|-\\frac{1}{24}\\text{ln}|{x}^{2}+2x+4|-\\frac{\\sqrt{3}}{12}{\\tan}^{-1}\\left(\\frac{x+1}{\\sqrt{3}}\\right)+C[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040774388\">Here again, we can drop the absolute value if we wish to do so, since [latex]{x}^{2}+2x+4&gt;0[\/latex] for all [latex]x[\/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<\/div>\r\n<div id=\"fs-id1165042311614\" data-type=\"example\">\r\n<div id=\"fs-id1165042088243\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042088245\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Finding a Volume<\/h3>\r\n<div id=\"fs-id1165042088245\" data-type=\"problem\">\r\n<p id=\"fs-id1165042088250\">Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of [latex]f\\left(x\\right)=\\frac{{x}^{2}}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[0,1\\right][\/latex] about the <em data-effect=\"italics\">y<\/em>-axis.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1165040716380\" data-type=\"solution\">\r\n<p id=\"fs-id1165040716383\">Let\u2019s begin by sketching the region to be revolved (see Figure 1). From the sketch, we see that the shell method is a good choice for solving this problem.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_04_001\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"276\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233822\/CNX_Calc_Figure_07_04_001.jpg\" alt=\"This figure is the graph of the function f(x) = x^2\/(x^2+1)^2. It is a curve above the x-axis. It is decreasing in the second quadrant, intersects at the origin, and increases in the first quadrant. Between x = 0 and x = 1, there is shaded area under the curve.\" width=\"276\" height=\"309\" data-media-type=\"image\/jpeg\" \/> Figure 1. We can use the shell method to find the volume of revolution obtained by revolving the region shown about the y-axis.[\/caption]<\/figure>\r\n<p id=\"fs-id1165042311737\">The volume is given by<\/p>\r\n\r\n<div id=\"fs-id1165042311740\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]V=2\\pi {\\displaystyle\\int }_{0}^{1}x\\cdot \\frac{{x}^{2}}{{\\left({x}^{2}+1\\right)}^{2}}dx=2\\pi {\\displaystyle\\int }_{0}^{1}\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165040669920\">Since [latex]\\text{deg}\\left({\\left({x}^{2}+1\\right)}^{2}\\right)=4&gt;3=\\text{deg}\\left({x}^{3}\\right)[\/latex], we can proceed with partial fraction decomposition. Note that [latex]{\\left({x}^{2}+1\\right)}^{2}[\/latex] is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get<\/p>\r\n\r\n<div id=\"fs-id1165041899092\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}=\\frac{Ax+B}{{x}^{2}+1}+\\frac{Cx+D}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042002713\">Finding a common denominator and equating the numerators gives<\/p>\r\n\r\n<div id=\"fs-id1165042002716\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{x}^{3}=\\left(Ax+B\\right)\\left({x}^{2}+1\\right)+Cx+D[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042272903\">Solving, we obtain [latex]A=1[\/latex], [latex]B=0[\/latex], [latex]C=-1[\/latex], and [latex]D=0[\/latex]. Substituting back into the integral, we have<\/p>\r\n\r\n<div id=\"fs-id1165040743214\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill V&amp; =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}dx\\hfill \\\\ &amp; =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}\\left(\\frac{x}{{x}^{2}+1}-\\frac{x}{{\\left({x}^{2}+1\\right)}^{2}}\\right)dx\\hfill \\\\ &amp; =2\\pi \\left(\\frac{1}{2}\\text{ln}\\left({x}^{2}+1\\right)+\\frac{1}{2}\\cdot \\frac{1}{{x}^{2}+1}\\right)|{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}1\\\\ \\end{array}}\\hfill \\\\ &amp; =\\pi \\left(\\text{ln}2-\\frac{1}{2}\\right).\\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-id1165042233003\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1165041926583\" data-type=\"exercise\">\r\n<div id=\"fs-id1165041926585\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1165041926585\" data-type=\"problem\">\r\n<p id=\"fs-id1165041926587\">Set up the partial fraction decomposition for [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{\\left(x+2\\right){\\left(x - 3\\right)}^{2}{\\left({x}^{2}+4\\right)}^{2}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558839\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558839\"]\r\n<div id=\"fs-id1165040744484\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1165040744492\">Use the problem-solving strategy.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558849\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1165041889366\" data-type=\"solution\">\r\n<p id=\"fs-id1165041889368\">[latex]\\frac{{x}^{2}+3x+1}{\\left(x+2\\right){\\left(x - 3\\right)}^{2}{\\left({x}^{2}+4\\right)}^{2}}=\\frac{A}{x+2}+\\frac{B}{x - 3}+\\frac{C}{{\\left(x - 3\\right)}^{2}}+\\frac{Dx+E}{{x}^{2}+4}+\\frac{Fx+G}{{\\left({x}^{2}+4\\right)}^{2}}[\/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=2068&amp;end=2110&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\/3.4PartialFractions2068to2110_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-id1165040744499\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize quadratic factors in a rational function<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1165042042804\" data-depth=\"1\">\n<h2>The General Method<\/h2>\n<p id=\"fs-id1165042042809\">Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let\u2019s outline the basic method in the following problem-solving strategy.<\/p>\n<div id=\"fs-id1165040639180\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Problem-Solving Strategy: Partial Fraction Decomposition<\/h3>\n<p id=\"fs-id1165040639187\">To decompose the rational function [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex], use the following steps:<\/p>\n<ol id=\"fs-id1165042108900\" type=\"1\">\n<li>Make sure that [latex]\\text{degree}\\left(P\\left(x\\right)\\right)<\\text{degree}\\left(Q\\left(x\\right)\\right)[\/latex]. If not, perform long division of polynomials.<\/li>\n<li>Factor [latex]Q\\left(x\\right)[\/latex] into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.<\/li>\n<li>Assuming that [latex]\\text{deg}\\left(P\\left(x\\right)\\right)<\\text{deg}\\left(Q\\left(x\\right)\\right)[\/latex], the factors of [latex]Q\\left(x\\right)[\/latex] determine the form of the decomposition of [latex]\\frac{P\\left(x\\right)}{Q\\left(x\\right)}[\/latex].\n\n\n<ol id=\"fs-id1165040671939\" type=\"a\">\n<li>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},...{A}_{n}[\/latex] satisfying<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165041899414\" 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;<\/li>\n<li>If [latex]Q\\left(x\\right)[\/latex] contains the repeated linear factor [latex]{\\left(ax+b\\right)}^{n}[\/latex], then the decomposition must contain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165040643994\" 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;<\/li>\n<li>For each irreducible quadratic factor [latex]a{x}^{2}+bx+c[\/latex] that [latex]Q\\left(x\\right)[\/latex] contains, the decomposition must include<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165042277015\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{Ax+B}{a{x}^{2}+bx+c}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>For each repeated irreducible quadratic factor [latex]{\\left(a{x}^{2}+bx+c\\right)}^{n}[\/latex], the decomposition must include<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165042230430\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{A}_{1}x+{B}_{1}}{a{x}^{2}+bx+c}+\\frac{{A}_{2}x+{B}_{2}}{{\\left(a{x}^{2}+bx+c\\right)}^{2}}+\\cdots +\\frac{{A}_{n}x+{B}_{n}}{{\\left(a{x}^{2}+bx+c\\right)}^{n}}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>After the appropriate decomposition is determined, solve for the constants.<\/li>\n<li>Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165040671182\" data-depth=\"1\">\n<h2 data-type=\"title\">Simple Quadratic Factors<\/h2>\n<p id=\"fs-id1165040671188\">Now let\u2019s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic [latex]a{x}^{2}+bx+c[\/latex] is irreducible if [latex]a{x}^{2}+bx+c=0[\/latex] has no real zeros\u2014that is, if [latex]{b}^{2}-4ac<0[\/latex].<\/p>\n<div id=\"fs-id1165042096550\" data-type=\"example\">\n<div id=\"fs-id1165042096552\" data-type=\"exercise\">\n<div id=\"fs-id1165042096554\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Rational Expressions with an Irreducible Quadratic Factor<\/h3>\n<div id=\"fs-id1165042096554\" data-type=\"problem\">\n<p id=\"fs-id1165042056695\">Evaluate [latex]\\displaystyle\\int \\frac{2x - 3}{{x}^{3}+x}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558879\">Show Solution<\/span><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040713993\" data-type=\"solution\">\n<p id=\"fs-id1165041966107\">Since [latex]\\text{deg}\\left(2x - 3\\right)<\\text{deg}\\left({x}^{3}+x\\right)[\/latex], factor the denominator and proceed with partial fraction decomposition. Since [latex]{x}^{3}+x=x\\left({x}^{2}+1\\right)[\/latex] contains the irreducible quadratic factor [latex]{x}^{2}+1[\/latex], include [latex]\\frac{Ax+B}{{x}^{2}+1}[\/latex] as part of the decomposition, along with [latex]\\frac{C}{x}[\/latex] for the linear term [latex]x[\/latex]. Thus, the decomposition has the form<\/p>\n<div id=\"fs-id1165042233088\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{2x - 3}{x\\left({x}^{2}+1\\right)}=\\frac{Ax+B}{{x}^{2}+1}+\\frac{C}{x}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040745299\">After getting a common denominator and equating the numerators, we obtain the equation<\/p>\n<div id=\"fs-id1165040745302\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2x - 3=\\left(Ax+B\\right)x+C\\left({x}^{2}+1\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041952135\">Solving for [latex]A,B[\/latex], and [latex]C[\/latex], we get [latex]A=3[\/latex], [latex]B=2[\/latex], and [latex]C=-3[\/latex].<\/p>\n<p id=\"fs-id1165042273867\">Thus,<\/p>\n<div id=\"fs-id1165042273871\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{2x - 3}{{x}^{3}+x}=\\frac{3x+2}{{x}^{2}+1}-\\frac{3}{x}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040757650\">Substituting back into the integral, we obtain<\/p>\n<div id=\"fs-id1165040757653\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{2x - 3}{{x}^{3}+x}dx}& ={\\displaystyle\\int \\left(\\frac{3x+2}{{x}^{2}+1}-\\frac{3}{x}\\right)dx}\\hfill & & & \\\\ & =3{\\displaystyle\\int \\frac{x}{{x}^{2}+1}dx+2\\displaystyle\\int \\frac{1}{{x}^{2}+1}dx - 3\\displaystyle\\int \\frac{1}{x}dx}\\hfill & & & \\text{Split up the integral.}\\hfill \\\\ & =\\frac{3}{2}\\text{ln}|{x}^{2}+1|+2{\\tan}^{-1}x - 3\\text{ln}|x|+C.\\hfill & & & \\text{Evaluate each integral.}\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040775841\"><em data-effect=\"italics\">Note<\/em>: We may rewrite [latex]\\text{ln}|{x}^{2}+1|=\\text{ln}\\left({x}^{2}+1\\right)[\/latex], if we wish to do so, since [latex]{x}^{2}+1>0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042107572\" data-type=\"example\">\n<div id=\"fs-id1165042107574\" data-type=\"exercise\">\n<div id=\"fs-id1165042107576\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Partial Fractions with an Irreducible Quadratic Factor<\/h3>\n<div id=\"fs-id1165042107576\" data-type=\"problem\">\n<p id=\"fs-id1165042107582\">Evaluate [latex]\\displaystyle\\int \\frac{dx}{{x}^{3}-8}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558869\">Show Solution<\/span><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041952824\" data-type=\"solution\">\n<p id=\"fs-id1165041952826\">We can start by factoring [latex]{x}^{3}-8=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]. We see that the quadratic factor [latex]{x}^{2}+2x+4[\/latex] is irreducible since [latex]{2}^{2}-4\\left(1\\right)\\left(4\\right)=-12<0[\/latex]. Using the decomposition described in the problem-solving strategy, we get<\/p>\n<div id=\"fs-id1165041874053\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)}=\\frac{A}{x - 2}+\\frac{Bx+C}{{x}^{2}+2x+4}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042238062\">After obtaining a common denominator and equating the numerators, this becomes<\/p>\n<div id=\"fs-id1165042238065\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1=A\\left({x}^{2}+2x+4\\right)+\\left(Bx+C\\right)\\left(x - 2\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041830086\">Applying either method, we get [latex]A=\\frac{1}{12},B=-\\frac{1}{12},\\text{and}C=-\\frac{1}{3}[\/latex].<\/p>\n<p id=\"fs-id1165042243673\">Rewriting [latex]\\displaystyle\\int \\frac{dx}{{x}^{3}-8}[\/latex], we have<\/p>\n<div id=\"fs-id1165042047720\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{dx}{{x}^{3}-8}=\\frac{1}{12}\\displaystyle\\int \\frac{1}{x - 2}dx-\\frac{1}{12}\\displaystyle\\int \\frac{x+4}{{x}^{2}+2x+4}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165041836432\">We can see that<\/p>\n<p id=\"fs-id1165041836436\">[latex]\\displaystyle\\int \\frac{1}{x - 2}dx=\\text{ln}|x - 2|+C[\/latex], but [latex]\\displaystyle\\int \\frac{x+4}{{x}^{2}+2x+4}dx[\/latex] requires a bit more effort. Let\u2019s begin by completing the square on [latex]{x}^{2}+2x+4[\/latex] to obtain<\/p>\n<div id=\"fs-id1165040772336\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{x}^{2}+2x+4={\\left(x+1\\right)}^{2}+3[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040638619\">By letting [latex]u=x+1[\/latex] and consequently [latex]du=dx[\/latex], we see that<\/p>\n<div id=\"fs-id1165042092238\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{x+4}{{x}^{2}+2x+4}dx}& ={\\displaystyle\\int \\frac{x+4}{{\\left(x+1\\right)}^{2}+3}dx}\\hfill & & & \\begin{array}{c}\\text{Complete the square on the}\\hfill \\\\ \\text{denominator.}\\hfill \\end{array}\\hfill \\\\ & ={\\displaystyle\\int \\frac{u+3}{{u}^{2}+3}du}\\hfill & & & \\begin{array}{c}\\text{Substitute}u=x+1,x=u - 1,\\hfill \\\\ \\text{and}du=dx.\\hfill \\end{array}\\hfill \\\\ & ={\\displaystyle\\int \\frac{u}{{u}^{2}+3}du+\\displaystyle\\int \\frac{3}{{u}^{2}+3}du}\\hfill & & & \\text{Split the numerator apart.}\\hfill \\\\ & =\\frac{1}{2}\\text{ln}|{u}^{2}+3|+\\frac{3}{\\sqrt{3}}{\\tan}^{-1}\\frac{u}{\\sqrt{3}}+C\\hfill & & & \\text{Evaluate each integral.}\\hfill \\\\ & =\\frac{1}{2}\\text{ln}|{x}^{2}+2x+4|+\\sqrt{3}{\\tan}^{-1}\\left(\\frac{x+1}{\\sqrt{3}}\\right)+C.\\hfill & & & \\begin{array}{c}\\text{Rewrite in terms of}x\\text{and}\\hfill \\\\ \\text{simplify.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040772126\">Substituting back into the original integral and simplifying gives<\/p>\n<div id=\"fs-id1165040772130\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int}\\frac{dx}{{x}^{3}-8}=\\frac{1}{12}\\text{ln}|x - 2|-\\frac{1}{24}\\text{ln}|{x}^{2}+2x+4|-\\frac{\\sqrt{3}}{12}{\\tan}^{-1}\\left(\\frac{x+1}{\\sqrt{3}}\\right)+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040774388\">Here again, we can drop the absolute value if we wish to do so, since [latex]{x}^{2}+2x+4>0[\/latex] for all [latex]x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042311614\" data-type=\"example\">\n<div id=\"fs-id1165042088243\" data-type=\"exercise\">\n<div id=\"fs-id1165042088245\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Finding a Volume<\/h3>\n<div id=\"fs-id1165042088245\" data-type=\"problem\">\n<p id=\"fs-id1165042088250\">Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of [latex]f\\left(x\\right)=\\frac{{x}^{2}}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[0,1\\right][\/latex] about the <em data-effect=\"italics\">y<\/em>-axis.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558859\">Show Solution<\/span><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040716380\" data-type=\"solution\">\n<p id=\"fs-id1165040716383\">Let\u2019s begin by sketching the region to be revolved (see Figure 1). From the sketch, we see that the shell method is a good choice for solving this problem.<\/p>\n<figure id=\"CNX_Calc_Figure_07_04_001\"><figcaption><\/figcaption><div style=\"width: 286px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233822\/CNX_Calc_Figure_07_04_001.jpg\" alt=\"This figure is the graph of the function f(x) = x^2\/(x^2+1)^2. It is a curve above the x-axis. It is decreasing in the second quadrant, intersects at the origin, and increases in the first quadrant. Between x = 0 and x = 1, there is shaded area under the curve.\" width=\"276\" height=\"309\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. We can use the shell method to find the volume of revolution obtained by revolving the region shown about the y-axis.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165042311737\">The volume is given by<\/p>\n<div id=\"fs-id1165042311740\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]V=2\\pi {\\displaystyle\\int }_{0}^{1}x\\cdot \\frac{{x}^{2}}{{\\left({x}^{2}+1\\right)}^{2}}dx=2\\pi {\\displaystyle\\int }_{0}^{1}\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165040669920\">Since [latex]\\text{deg}\\left({\\left({x}^{2}+1\\right)}^{2}\\right)=4>3=\\text{deg}\\left({x}^{3}\\right)[\/latex], we can proceed with partial fraction decomposition. Note that [latex]{\\left({x}^{2}+1\\right)}^{2}[\/latex] is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get<\/p>\n<div id=\"fs-id1165041899092\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}=\\frac{Ax+B}{{x}^{2}+1}+\\frac{Cx+D}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042002713\">Finding a common denominator and equating the numerators gives<\/p>\n<div id=\"fs-id1165042002716\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{x}^{3}=\\left(Ax+B\\right)\\left({x}^{2}+1\\right)+Cx+D[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042272903\">Solving, we obtain [latex]A=1[\/latex], [latex]B=0[\/latex], [latex]C=-1[\/latex], and [latex]D=0[\/latex]. Substituting back into the integral, we have<\/p>\n<div id=\"fs-id1165040743214\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill V& =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}dx\\hfill \\\\ & =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}\\left(\\frac{x}{{x}^{2}+1}-\\frac{x}{{\\left({x}^{2}+1\\right)}^{2}}\\right)dx\\hfill \\\\ & =2\\pi \\left(\\frac{1}{2}\\text{ln}\\left({x}^{2}+1\\right)+\\frac{1}{2}\\cdot \\frac{1}{{x}^{2}+1}\\right)|{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}1\\\\ \\end{array}}\\hfill \\\\ & =\\pi \\left(\\text{ln}2-\\frac{1}{2}\\right).\\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-id1165042233003\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1165041926583\" data-type=\"exercise\">\n<div id=\"fs-id1165041926585\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1165041926585\" data-type=\"problem\">\n<p id=\"fs-id1165041926587\">Set up the partial fraction decomposition for [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{\\left(x+2\\right){\\left(x - 3\\right)}^{2}{\\left({x}^{2}+4\\right)}^{2}}dx[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558839\">Hint<\/span><\/p>\n<div id=\"q44558839\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165040744484\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1165040744492\">Use the problem-solving strategy.<\/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=\"q44558849\">Show Solution<\/span><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165041889366\" data-type=\"solution\">\n<p id=\"fs-id1165041889368\">[latex]\\frac{{x}^{2}+3x+1}{\\left(x+2\\right){\\left(x - 3\\right)}^{2}{\\left({x}^{2}+4\\right)}^{2}}=\\frac{A}{x+2}+\\frac{B}{x - 3}+\\frac{C}{{\\left(x - 3\\right)}^{2}}+\\frac{Dx+E}{{x}^{2}+4}+\\frac{Fx+G}{{\\left({x}^{2}+4\\right)}^{2}}[\/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=2068&amp;end=2110&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\/3.4PartialFractions2068to2110_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-id1165040744499\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\"><\/div>\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-1587\">\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":14,"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-1587","chapter","type-chapter","status-publish","hentry"],"part":158,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1587","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":6,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1587\/revisions"}],"predecessor-version":[{"id":2429,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1587\/revisions\/2429"}],"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\/1587\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1587"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1587"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1587"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1587"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}