# How do you use partial fractions to find the integral int (e^x)/((e^x-1)(e^x+4))dx?

Jul 20, 2017

$\int {e}^{x} / \left(\left({e}^{x} - 1\right) \left({e}^{x} + 4\right)\right) \mathrm{dx} = \frac{1}{5} \left(\ln | 1 - {e}^{- x} | - \ln | 1 + 4 {e}^{- x} |\right)$

#### Explanation:

Let ${e}^{x} / \left(\left({e}^{x} - 1\right) \left({e}^{x} + 4\right)\right) = \frac{A}{{e}^{x} - 1} + \frac{B}{{e}^{x} + 4}$

$= \frac{A {e}^{x} + 4 A + B {e}^{x} - B}{\left({e}^{x} - 1\right) \left({e}^{x} + 4\right)} = \frac{{e}^{x} \left(A + B\right) + 4 A - B}{\left({e}^{x} - 1\right) \left({e}^{x} + 4\right)}$

Comparing like terms, $4 A - B = 0$ i.e. $B = 4 A$ and $A + B = 1$

i.e. $A = \frac{1}{5}$ and $B = \frac{4}{5}$ and

$\int {e}^{x} / \left(\left({e}^{x} - 1\right) \left({e}^{x} + 4\right)\right) \mathrm{dx} = \frac{1}{5} \int \frac{1}{{e}^{x} - 1} \mathrm{dx} + \frac{4}{5} \int \frac{1}{{e}^{x} + 4} \mathrm{dx}$

= $\frac{1}{5} \int {e}^{- x} / \left(1 - {e}^{- x}\right) \mathrm{dx} + \frac{4}{5} \int {e}^{- x} / \left(1 + 4 {e}^{- x}\right) \mathrm{dx}$

Let $u = 1 - {e}^{- x}$ then $\mathrm{du} = {e}^{- x} \mathrm{dx}$ and

$v = 1 + 4 {e}^{- x}$ then $\mathrm{dv} = - 4 {e}^{- x} \mathrm{dx}$

Hence $\frac{1}{5} \int {e}^{- x} / \left(1 - {e}^{- x}\right) \mathrm{dx} + \frac{4}{5} \int {e}^{- x} / \left(1 + 4 {e}^{- x}\right) \mathrm{dx}$

= $\frac{1}{5} \int \frac{\mathrm{du}}{u} - \frac{1}{5} \int \frac{\mathrm{dv}}{v}$

= $\frac{1}{5} \left(\ln | u | - \ln | v |\right)$

= $\frac{1}{5} \left(\ln | 1 - {e}^{- x} | - \ln | 1 + 4 {e}^{- x} |\right)$