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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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  • (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,
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$


answered Feb 6, 2013 by meena.p
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