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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\frac{(x-3)e^x}{(x-1)^3}\]

$\begin{array}{1 1} \frac{e^x}{(x-1)^2}+c \\ \frac{e^x}{(1-x)^2}+c \\ \frac{e^x}{(x+1)^2}+c \\ \frac{e^x}{(x-3)^2}+c \end{array} $

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1 Answer

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Toolbox:
  • (i)$ \int e^x[f(x)+f'(x)]dx=e^x f(x)+c$
  • (ii) $ \frac{d}{dx}(\frac{1}{(x-1)^2})=-\frac{2}{(x-1)^3}$
$I= \int \frac {(x-3)}{(x-1)^3}e^x dx$
 
This can be written as
 
$\int \frac{(x-1-2)}{(x-1)^3} e^x dx.$
 
On seperating this we get,
 
$=\int e^x\bigg[\frac{x-1}{(x-1)^3}-\frac{2}{(x-1)^3}\bigg]dx$
$=\int e^x\bigg[\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\bigg]dx$
 
Let $f(x)=\frac{1}{(x-1)^2}=(x-1)^{-2}$
 
on differentiating w.r.t.x we get,
 
$=-2(x-1)^3=\frac{-2}{(x-1)^3}$
 
clearly this is of the form
 
$\int [f(x)+f'(x)]=e^x[f(x)]+c.$
 
Hence $ I=\int e^x\bigg[\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\bigg]dx$
 
$=e^x\bigg[\frac{1}{(x-1)^2}\bigg]+c$

 

answered Feb 6, 2013 by meena.p
 
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