Integrals Involving Logarithmic Functions

Learning Outcomes

Integrate functions involving logarithmic functions

Integrating functions of the form $f(x)={x}^{-1}$ result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as $f(x)=\text{ln}x$ and $f(x)={\text{log}}_{a}x,$ are also included in the rule.

Integration Formulas Involving Logarithmic Functions

The following formulas can be used to evaluate integrals involving logarithmic functions.

\begin{matrix} \int x^{- 1} d x & = & ln | x | + C \int ln x d x & = & x ln x - x + C = x (ln x - 1) + C \int {log}_{a} x d x & = & \frac{x}{ln a} (ln x - 1) + C \end{matrix}

$\begin{array}{ccc}\hfill \displaystyle\int {x}^{-1}dx& =\hfill & \text{ln}|x|+C\hfill \\ \hfill \displaystyle\int \text{ln}xdx& =\hfill & x\text{ln}x-x+C=x(\text{ln}x-1)+C\hfill \\ \hfill \displaystyle\int {\text{log}}_{a}xdx& =\hfill & \frac{x}{\text{ln}a}(\text{ln}x-1)+C\hfill \end{array}$

Example: Finding an Antiderivative Involving $\text{ln}x$

Find the antiderivative of the function $\dfrac{3}{x-10}.$

Show Solution

First factor the 3 outside the integral symbol. Then use the $u^{-1}$ rule. Thus,

\begin{matrix} \int \frac{3}{x - 10} d x & = 3 \int \frac{1}{x - 10} d x = 3 \int \frac{d u}{u} = 3 ln | u | + C = 3 ln | x - 10 | + C, x \neq 10. \end{matrix}

$\begin{array}{ll}{\displaystyle\int \frac{3}{x-10}dx}\hfill & =3{\displaystyle\int \frac{1}{x-10}dx}\hfill \\ \\ \\ & =3{\displaystyle\int \frac{du}{u}}\hfill \\ & =3\text{ln}|u|+C\hfill \\ & =3\text{ln}|x-10|+C,x\ne 10.\hfill \end{array}$

See Figure 3.

A graph of the function f(x) = 3 / (x – 10). There is an asymptote at x=10. The first segment is a decreasing concave down curve that approaches 0 as x goes to negative infinity and approaches negative infinity as x goes to 10. The second segment is a decreasing concave up curve that approaches infinity as x goes to 10 and approaches 0 as x approaches infinity.

Figure 3. The domain of this function is $x\ne 10.$

Try It

Find the antiderivative of $\dfrac{1}{x+2}.$

Show Solution

$\text{ln}|x+2|+C$

Hint

Follow the pattern from the last example to solve the problem.

Example: Finding an Antiderivative of a Rational Function

Find the antiderivative of $\dfrac{2{x}^{3}+3x}{{x}^{4}+3{x}^{2}}.$

Show Solution

This can be rewritten as

\int (2 x^{3} + 3 x) {(x^{4} + 3 x^{2})}^{- 1} d x .

$\displaystyle\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx.$ Use substitution. Let

u = x^{4} + 3 x^{2},

$u={x}^{4}+3{x}^{2},$ then

d u = 4 x^{3} + 6 x .

$du=4{x}^{3}+6x.$ Alter du by factoring out the 2. Thus,

\begin{matrix} d u & = & (4 x^{3} + 6 x) d x = & 2 (2 x^{3} + 3 x) d x \frac{1}{2} d u & = & (2 x^{3} + 3 x) d x . \end{matrix}

$\begin{array}{}\\ \hfill du& =\hfill & (4{x}^{3}+6x)dx\hfill \\ & =\hfill & 2(2{x}^{3}+3x)dx\hfill \\ \hfill \frac{1}{2}du& =\hfill & (2{x}^{3}+3x)dx.\hfill \end{array}$

Rewrite the integrand in $u$ :

\int (2 x^{3} + 3 x) {(x^{4} + 3 x^{2})}^{- 1} d x = \frac{1}{2} \int u^{- 1} d u .

$\displaystyle\int (2{x}^{3}+3x){({x}^{4}+3{x}^{2})}^{-1}dx=\frac{1}{2}\displaystyle\int {u}^{-1}du.$

Then we have

\begin{matrix} \frac{1}{2} \int u^{- 1} d u & = \frac{1}{2} ln | u | + C = \frac{1}{2} ln | x^{4} + 3 x^{2} | + C . \end{matrix}

$\begin{array}{ll}\frac{1}{2}{\displaystyle\int {u}^{-1}du}\hfill & =\frac{1}{2}\text{ln}|u|+C\hfill \\ \\ & =\frac{1}{2}\text{ln}|{x}^{4}+3{x}^{2}|+C.\hfill \end{array}$

Watch the following video to see the worked solution to Example: Finding an Antiderivative of a Rational Function.

Closed Captioning and Transcript Information for Video

Example: Finding an Antiderivative of a Logarithmic Function

Find the antiderivative of the log function ${\text{log}}_{2}x.$

Show Solution

Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based on this format, we have

\int {log}_{2} x d x = \frac{x}{ln 2} (ln x - 1) + C .

$\displaystyle\int {\text{log}}_{2}xdx=\frac{x}{\text{ln}2}(\text{ln}x-1)+C.$

Try It

Find the antiderivative of ${\text{log}}_{3}x.$

Show Solution

$\frac{x}{\text{ln}3}(\text{ln}x-1)+C$

Hint

Follow the previous example and refer to the rule on integration formulas involving logarithmic functions.

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

The example below is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key to a smooth integration.

Evaluating a Definite Integral

Find the definite integral of ${\displaystyle\int }_{0}^{\pi \text{/}2}\frac{ \sin x}{1+ \cos x}dx.$

Show Solution

We need substitution to evaluate this problem. Let $u=1+ \cos x,,$ so $du=\text{−} \sin xdx.$ Rewrite the integral in terms of $u$ , changing the limits of integration as well. Thus,

\begin{matrix} u = 1 + cos (0) = 2 u = 1 + cos (\frac{π}{2}) = 1 \end{matrix}

$\begin{array}{c}u=1+ \cos (0)=2\hfill \\ u=1+ \cos (\frac{\pi }{2})=1\hfill \end{array}$

Then

\begin{matrix} \int_{0}^{π / 2} \frac{sin x}{1 + cos x} & = - \int_{2}^{1} u^{- 1} d u = \int_{1}^{2} u^{- 1} d u = {ln | u | |}_{1}^{2} = [ln 2 - ln 1] = ln 2 \end{matrix}

$\begin{array}{cc}{\displaystyle\int} _{0}^{\pi \text{/}2}\dfrac{ \sin x}{1+ \cos x}\hfill & =\text{−}{\displaystyle\int }_{2}^{1}{u}^{-1}du\hfill \\ \\ \\ & ={\displaystyle\int }_{1}^{2}{u}^{-1}du\hfill \\ & ={\text{ln}|u||}_{1}^{2}\hfill \\ & =\left[\text{ln}2-\text{ln}1\right]\hfill \\ & =\text{ln}2\hfill \end{array}$

Module 5: Integration