# How do you integrate int e^x/(7+e^(2x)) by trigonometric substitution?

Mar 24, 2018

$\frac{1}{\sqrt{7}} \arctan \left({e}^{x} / \sqrt{7}\right)$

#### Explanation:

$\int {e}^{x} / \left(7 + {e}^{2} x\right) \mathrm{dx} = \int {e}^{x} / \left(7 + {\left({e}^{x}\right)}^{2}\right) \mathrm{dx}$

write it in a format that we can look to integrate more easily

Let $u = {e}^{x}$
$\mathrm{du} = {e}^{x} \mathrm{dx}$

The integral in terms of u is now

$\int \frac{1}{7 + {u}^{2}} \cdot \mathrm{du}$

To use a trig substitution on this I used the fact that $\int \frac{1}{{a}^{2} + {x}^{2}} \mathrm{dx} = \frac{1}{a} \arctan \left(\frac{x}{a}\right)$

Therefore a good trig substitution for this sort of problem would be
to let $u = \sqrt{7} \tan \left(A\right)$ therefore ${u}^{2} = 7 {\tan}^{2} \left(A\right)$

$u = \sqrt{7} \tan A$

$u = \sqrt{7} S \in \frac{A}{C} o s \left(A\right)$

differentiating using the quotient rule

$\mathrm{du} = \sqrt{7} \left(\frac{\cos \left(A\right) \cos \left(A\right) - \sin \left(A\right) \left(- \sin \left(A\right)\right)}{\cos} ^ 2\right) \mathrm{dA}$

$N o t e {\cos}^{2} \left(A\right) + {\sin}^{2} \left(A\right) = 1$

$\mathrm{du} = \sqrt{7} \frac{1}{\cos} ^ 2 \left(A\right) \mathrm{dA}$

The integral is now of the form

$\int \frac{1}{7} \left(1 + {\tan}^{2} \left(A\right)\right) . \sqrt{7} \frac{1}{\cos} ^ 2 \left(A\right) \mathrm{dA}$

$= \int \frac{\sqrt{7}}{7} \left(1 + {\sin}^{2} \frac{A}{\cos} ^ 2 \left(A\right)\right) . \sqrt{7} \frac{1}{\cos} ^ 2 \left(A\right) \mathrm{dA}$

$= \int \frac{\sqrt{7}}{7} \cdot \frac{1}{\frac{{\cos}^{2} \left(A\right) + {\sin}^{2} \left(A\right)}{\cos} ^ 2 \left(A\right)} \frac{1}{\cos} ^ 2 \left(A\right) \mathrm{dA}$

$= \int \frac{\sqrt{7}}{7} \cdot 1 \mathrm{dA}$ (everything cancels out)

$= \frac{\sqrt{7}}{7} A$

Now remember
$u = \sqrt{7} \tan \left(A\right)$ and ${u}^{2} = 7 {\tan}^{2} \left(A\right)$

Therefore $\tan \left(A\right) = \frac{u}{\sqrt{7}}$

A =$\arctan \left(\frac{u}{\sqrt{7}}\right)$

So the integral results in = $\frac{\sqrt{7}}{7} \arctan \left(\frac{u}{\sqrt{7}}\right)$

= $\frac{1}{\sqrt{7}} \arctan \left(\frac{u}{\sqrt{7}}\right)$

finally we had $u = {e}^{x}$

Therefore the answer is $\frac{1}{\sqrt{7}} \arctan \left({e}^{x} / \sqrt{7}\right)$