# How do you use partial fractions to find the integral int (e^x)/((e^(2x)+1)(e^x-1))dx?

Mar 5, 2017

The integral equals $- \frac{1}{2} \arctan \left({e}^{x}\right) - \frac{1}{2} \ln \left(\frac{{e}^{x} + 1}{|} {e}^{x} - 1 |\right) + C$

#### Explanation:

First of all, the integral can be rewritten as

$\int {e}^{x} / \left(\left({\left({e}^{x}\right)}^{2} + 1\right) \left({e}^{x} - 1\right)\right) \mathrm{dx}$

Now let $u = {e}^{x}$. Then $\mathrm{du} = {e}^{x} \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{e} ^ x$.

$\int {e}^{x} / \left(\left({u}^{2} + 1\right) \left(u - 1\right)\right) \cdot \frac{\mathrm{du}}{e} ^ x$

$\int \frac{1}{\left({u}^{2} + 1\right) \left(u - 1\right)} \mathrm{du}$

Now use partial fractions.

$\frac{A u + B}{{u}^{2} + 1} + \frac{C}{u - 1} = \frac{1}{\left({u}^{2} + 1\right) \left(u - 1\right)}$

$\left(A u + B\right) \left(u - 1\right) + C \left({u}^{2} + 1\right) = 1$

$A {u}^{2} + B u - B - A u + C {u}^{2} + C = 1$

$\left(A + C\right) {u}^{2} + \left(B - A\right) u + \left(C - B\right) = 1$

We now write a system of equations:

$\left\{\begin{matrix}A + C = 0 \\ B - A = 0 \\ C - B = 1\end{matrix}\right.$

We can simplify to $C = - A$ and $B = A$ and substitute into the third equation.

$- A - A = 1$

$- 2 A = 1$

$A = - \frac{1}{2}$

This implies that $C = \frac{1}{2}$ and $B = - \frac{1}{2}$.

Our integral becomes:

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

$\int \frac{- \frac{1}{2} \left(u + 1\right)}{{u}^{2} + 1} + \frac{1}{2 \left(u - 1\right)} \mathrm{du}$

$\int - \frac{1}{2} \frac{u + 1}{{u}^{2} + 1} + \frac{1}{2 \left(u - 1\right)} \mathrm{du}$

$- \frac{1}{2} \int \frac{u + 1}{{u}^{2} + 1} \mathrm{du} + \int \frac{1}{2 \left(u - 1\right)} \mathrm{du}$

$- \frac{1}{2} \int \frac{u}{{u}^{2} + 1} + \frac{1}{{u}^{2} + 1} \mathrm{du} + \int \frac{1}{2 \left(u - 1\right)} \mathrm{du}$

$- \frac{1}{2} \arctan u - \frac{1}{2} \ln \left({u}^{2} + 1\right) + \frac{1}{2} \ln | u - 1 | + C$

$- \frac{1}{2} \arctan \left({e}^{x}\right) - \frac{1}{2} \ln \left({e}^{x} + 1\right) + \frac{1}{2} \ln | {e}^{x} - 1 | + C$

$- \frac{1}{2} \arctan \left({e}^{x}\right) - \frac{1}{2} \left(\ln \left({e}^{x} + 1\right) - \ln | {e}^{x} - 1 |\right) + C$

$- \frac{1}{2} \arctan \left({e}^{x}\right) - \frac{1}{2} \ln \left(\frac{{e}^{x} + 1}{|} {e}^{x} - 1 |\right) + C$

Practice Exercises

Solve the following integrals using partial fractions. You may use a u-substitution to begin.

a) $\int \frac{4 x + 1}{4 {x}^{2} + 12 x + 9} \mathrm{dx}$

b) $\int \sin \frac{x}{\cos x \left(\sin x - 1\right)} \mathrm{dx}$

Solutions

a) $\ln | 2 x + 3 | + \frac{5}{4 x + 6} + C$
b) $\frac{1}{4} \ln | \frac{\sin x + 1}{1 - \sin x} | + \frac{1}{2 \left(\sin x - 1\right)} + C$

Hopefully this helps!