# 5.6 Integrals involving exponential and logarithmic functions  (Page 3/4)

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Suppose the rate of growth of the fly population is given by $g\left(t\right)={e}^{0.01t},$ and the initial fly population is 100 flies. How many flies are in the population after 15 days?

There are 116 flies.

## Evaluating a definite integral using substitution

Evaluate the definite integral using substitution: ${\int }_{1}^{2}\frac{{e}^{1\text{/}x}}{{x}^{2}}dx.$

This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as a power of x , then bring the x 2 in the denominator up to the numerator using a negative exponent. We have

${\int }_{1}^{2}\frac{{e}^{1\text{/}x}}{{x}^{2}}dx={\int }_{1}^{2}{e}^{{x}^{-1}}{x}^{-2}dx.$

Let $u={x}^{-1},$ the exponent on e . Then

$\begin{array}{cc}\hfill du& =\text{−}{x}^{-2}dx\hfill \\ \hfill -du& ={x}^{-2}dx.\hfill \end{array}$

Bringing the negative sign outside the integral sign, the problem now reads

$\text{−}\int {e}^{u}du.$

Next, change the limits of integration:

$\begin{array}{}\\ \\ u={\left(1\right)}^{-1}=1\hfill \\ u={\left(2\right)}^{-1}=\frac{1}{2}.\hfill \end{array}$

Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange the limits. Thus,

$\begin{array}{}\\ \\ \\ \text{−}{\int }_{1}^{1\text{/}2}{e}^{u}du\hfill & ={\int }_{1\text{/}2}^{1}{e}^{u}du\hfill \\ & ={e}^{u}{|}_{1\text{/}2}^{1}\hfill \\ & =e-{e}^{1\text{/}2}\hfill \\ & =e-\sqrt{e}.\hfill \end{array}$

Evaluate the definite integral using substitution: ${\int }_{1}^{2}\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx.$

${\int }_{1}^{2}\frac{1}{{x}^{3}}{e}^{4{x}^{-2}}dx=\frac{1}{8}\left[{e}^{4}-e\right]$

## Integrals involving logarithmic functions

Integrating functions of the form $f\left(x\right)={x}^{-1}$ result in the absolute value of the natural log function, as shown in the following rule. Integral formulas for other logarithmic functions, such as $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}x$ and $f\left(x\right)={\text{log}}_{a}x,$ are also included in the rule.

## Rule: integration formulas involving logarithmic functions

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

$\begin{array}{ccc}\hfill \int {x}^{-1}dx& =\hfill & \text{ln}|x|+C\hfill \\ \hfill \int \text{ln}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx& =\hfill & x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x-x+C=x\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x-1\right)+C\hfill \\ \hfill \int {\text{log}}_{a}x\phantom{\rule{0.2em}{0ex}}dx& =\hfill & \frac{x}{\text{ln}\phantom{\rule{0.1em}{0ex}}a}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x-1\right)+C\hfill \end{array}$

## Finding an antiderivative involving $\text{ln}\phantom{\rule{0.2em}{0ex}}x$

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

First factor the 3 outside the integral symbol. Then use the u −1 rule. Thus,

$\begin{array}{ll}\int \frac{3}{x-10}dx\hfill & =3\int \frac{1}{x-10}dx\hfill \\ \\ \\ & =3\int \frac{du}{u}\hfill \\ & =3\phantom{\rule{0.1em}{0ex}}\text{ln}|u|+C\hfill \\ & =3\phantom{\rule{0.1em}{0ex}}\text{ln}|x-10|+C,x\ne 10.\hfill \end{array}$

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

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

## Finding an antiderivative of a rational function

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

This can be rewritten as $\int \left(2{x}^{3}+3x\right){\left({x}^{4}+3{x}^{2}\right)}^{-1}dx.$ Use substitution. Let $u={x}^{4}+3{x}^{2},$ then $du=4{x}^{3}+6x.$ Alter du by factoring out the 2. Thus,

$\begin{array}{}\\ \hfill du& =\hfill & \left(4{x}^{3}+6x\right)dx\hfill \\ & =\hfill & 2\left(2{x}^{3}+3x\right)dx\hfill \\ \hfill \frac{1}{2}\phantom{\rule{0.2em}{0ex}}du& =\hfill & \left(2{x}^{3}+3x\right)dx.\hfill \end{array}$

Rewrite the integrand in u :

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

Then we have

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

## Finding an antiderivative of a logarithmic function

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

$\int {\text{log}}_{2}xdx=\frac{x}{\text{ln}\phantom{\rule{0.1em}{0ex}}2}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x-1\right)+C.$

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

$\frac{x}{\text{ln}\phantom{\rule{0.1em}{0ex}}3}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x-1\right)+C$

[link] 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 ${\int }_{0}^{\pi \text{/}2}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{1+\text{cos}\phantom{\rule{0.1em}{0ex}}x}dx.$

We need substitution to evaluate this problem. Let $u=1+\text{cos}\phantom{\rule{0.1em}{0ex}}x,,$ so $du=\text{−}\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}dx.$ Rewrite the integral in terms of u , changing the limits of integration as well. Thus,

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

Then

$\begin{array}{cc}{\int }_{0}^{\pi \text{/}2}\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{1+\text{cos}\phantom{\rule{0.1em}{0ex}}x}\hfill & =\text{−}{\int }_{2}^{1}{u}^{-1}du\hfill \\ \\ \\ & ={\int }_{1}^{2}{u}^{-1}du\hfill \\ & ={\text{ln}|u||}_{1}^{2}\hfill \\ & =\left[\text{ln}\phantom{\rule{0.1em}{0ex}}2-\text{ln}\phantom{\rule{0.1em}{0ex}}1\right]\hfill \\ & =\text{ln}\phantom{\rule{0.1em}{0ex}}2.\hfill \end{array}$

## Key concepts

• Exponential and logarithmic functions arise in many real-world applications, especially those involving growth and decay.
• Substitution is often used to evaluate integrals involving exponential functions or logarithms.

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
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