# How do you integrate int e^x/[(e^x-2)(e^(2x)+1)]dx using partial fractions?

Nov 10, 2016

$\frac{1}{5} \left(\ln \left\mid {e}^{x} - 2 \right\mid - \frac{1}{2} \ln \left\mid {e}^{2 x} + 1 \right\mid - 2 \arctan {e}^{x}\right) + c$

#### Explanation:

Substitution with ${e}^{x} = y$ then partial fractions

$x = \ln y \setminus \setminus , \setminus \setminus \mathrm{dx} = \frac{1}{y} \mathrm{dy}$

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

Search $A , B , C$ so that

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

So $A = - B$, $C = 2 B$ and $A - 2 C = - B - 4 B = 1$

Then $B = - \frac{1}{5} , A = \frac{1}{5} , C = - \frac{2}{5}$

$\int \frac{1}{\left(y - 2\right) \left({y}^{2} + 1\right)} \mathrm{dy} =$
$= \frac{1}{5} \left(\int \frac{1}{y - 2} \mathrm{dy} - \frac{1}{2} \int \frac{2 y}{{y}^{2} + 1} \mathrm{dy} - 2 \int \frac{1}{{y}^{2} + 1} \mathrm{dy}\right) =$
$= \frac{1}{5} \left(\ln \left\mid y - 2 \right\mid - \frac{1}{2} \ln \left\mid {y}^{2} + 1 \right\mid - 2 \arctan y\right) + c$

Then substitute $y$ with ${e}^{x}$