# Integrate the rational functions$\frac{1}{(e^x-1)} [Hint:Put\;e^x=\;t]$

$\begin{array}{1 1} \log \large | \frac{(e^x-1)}{(e^x)} | +c \\ \log \large | \frac{(e^x-1)}{(e^x+1)} | +c \\ \log \large | \frac{(e^x+1)}{(e^x)} | +c \\ \log \large | \frac{(e^x+1)}{(e^x-1)} | +c \end{array}$

Toolbox:
• $\;$Form of the rational function is $\frac{px+q}{(x+a)(x+b)}$,then it can be resolved into its partial function as $\frac{A}{(x+a)}+\frac{B}{(x+b)}.$
• $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
Given $I=\int\frac{1}{e^x-1}dx.$

Let $e^x=t.$

On differentiating we get,

$e^xdx=dt.$

$\Rightarrow dx=\frac{dt}{t}.$($e^x=t.$)

Hence $\int\frac{dx}{e^x-1}=\int\frac{dt/t}{t-1}=\int\frac{dt}{t(t-1)}.$

Resolve into partial fraction,

$\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}.$

$1=A(t-1)+Bt.$

Equating the coefficients of t we get

0=A+B---(1)

Equating the constant terms,

1=-A$\Rightarrow A=-1.$

Substituting for a in equ(1) we get,

-1+B=0$\Rightarrow B=1.$

Hence A=-1 and B=1

on substituting for A and B,

Therefore $\frac{1}{t(t+1)}=\frac{-1}{t}+\frac{1}{t-1}.$

$I=\int-\frac{dt}{t}+\int\frac{dt}{t-1}.$

On integrating we get,

$\;\;\;=-log|t|+log|t-1|+c.$

log a-log b=$log|\frac{a}{b}|$,similarly

$\;\;\;=log\frac|t-1|{t}+c.$

Substituting for t,

$\;\;\;=log\frac{(e^x-1)}{(e^x)}+c.$

edited Jul 21, 2013