Integrals Involving Exponential and Logarithmic Functions: Learn It 2

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Integrals Involving Exponential and Logarithmic Functions: Learn It 2

Integrals 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{array}{ccc} \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{array}

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

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

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

\begin{array}{ll} \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{array}

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

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 $x\ne 10.$

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

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 $u={x}^{4}+3{x}^{2},$ then $du=4{x}^{3}+6x.$ 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{array}{ll} \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{array}

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

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “5.6.2” here (opens in new window).

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

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 .

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

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.

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

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{array}{cc} \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{array}

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