# How do you evaluate the integral of [ (e^7x) / ((e^14x)+9) dx ]?

May 4, 2015

$\int \left\{\frac{a x}{b x + c}\right\} \mathrm{dx} = \frac{a c}{b} ^ 2 \left(\frac{b}{c} x - \ln | \frac{b}{c} x + 1 |\right) + C$

where $a$, $b$, and $c$ are constants. $C$ is the constant of integration.
I got the above result by the following process:

$\int \left\{\frac{a x}{b x + c}\right\} \mathrm{dx} = \int \left\{\frac{a}{b} \frac{\frac{b}{c} x}{\frac{b}{c} x + 1}\right\} \mathrm{dx}$

Let $u = \frac{b}{c} x$, Then $\mathrm{dx} = \frac{c}{b} \mathrm{du}$.

By substituting, the integral in terms of $u$ is,

$\frac{a c}{b} ^ 2 \int \left\{\frac{u}{u + 1}\right\} \mathrm{du} = \frac{a c}{b} ^ 2 \int \left\{1 - \frac{1}{u + 1}\right\} \mathrm{du}$

$= \frac{a c}{b} ^ 2 \left(u - \ln | u + 1 |\right) + C$

Now that the integration is done, express the result in terms of $x$. Substitute $u = \frac{b}{c} x$ and get

$= \frac{a c}{b} ^ 2 \left(\frac{b}{c} x - \ln | \frac{b}{c} x + 1 |\right) + C$

May 13, 2015

I have another method

Substitute $u = {e}^{7 x}$

$\mathrm{du} = 7 {e}^{7 x} \mathrm{dx}$

$\int \frac{{e}^{7 x}}{{e}^{14 x} + 9} \mathrm{dx} = \frac{1}{7} \int \frac{7 {e}^{7 x}}{{e}^{14 x} + 9} \mathrm{dx}$

${u}^{2} = {e}^{14 x}$

$\frac{1}{7} \int \frac{1}{{u}^{2} + 9} \mathrm{du} = \frac{1}{63} \int \frac{1}{\frac{1}{9} {u}^{2} + 1} \mathrm{du}$

Substitute again, $t = \frac{1}{3} u$ ${t}^{2} = \frac{1}{9} {u}^{2}$

$\frac{1}{21} \int \frac{1}{{t}^{2} + 1} \mathrm{dt}$

$\frac{1}{21} \left[\arctan \left(t\right)\right] + C$

Substitute back

$\frac{1}{21} \arctan \left(\frac{1}{3} {e}^{7} x\right) + C$