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Home  >>  CBSE XII  >>  Math  >>  Integrals
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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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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,
 
$\;\;\;=log\frac{(e^x-1)}{(e^x)}+c.$

 

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

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