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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} $

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  • $\;$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,
$\Rightarrow dx=\frac{dt}{t}.$($e^x=t.$)
Hence $\int\frac{dx}{e^x-1}=\int\frac{dt/t}{t-1}=\int\frac{dt}{t(t-1)}.$
Resolve into partial fraction,
Equating the coefficients of t we get
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 $\frac{1}{t(t+1)}=\frac{-1}{t}+\frac{1}{t-1}.$
On integrating we get,
log a-log b=$log|\frac{a}{b}|$,similarly
Substituting for t,


answered Feb 7, 2013 by sreemathi.v
edited Jul 21, 2013 by vijayalakshmi_ramakrishnans

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