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Find:    \int \frac{x-3}{(x-1)^3}e^xdx

 

 

 

 
 
 
 
 

Answers (1)

I= \int \frac{x-3}{(x-1)^3}e^xdx 

I=\int \frac{x-2-1}{(x-1)^3}dx\cdot e^x

I=\int \frac{x-1-2}{(x-1)^3} e^xdx\Rightarrow \int \left [ \frac{x-1}{(x-1)^3}-\frac{2}{(x-1)^3} \right ]e^xdx

I=\int \left [ \frac{1}{(x-1)^2}-\frac{2}{(x-1)^3} \right ]e^xdx     -----(1)

here,  \frac{\mathrm{d} }{\mathrm{d} x}\left [ \frac{1}{(x-1)^2} \right ]=\frac{\mathrm{d} }{\mathrm{d} x}(x-1)^{-2}=-2(x-1)^{-3}

Then, f'(x)=\frac{-2}{(x-1)^3}

We know that,

\int (f(x)+f'(x))e^xdx=e^xf(x)+c

Then from equation (1), we hav

\int \left [ \frac{1}{(x-1)^2}-\frac{2}{(x-1)^3} \right ]e^xdx=\frac{e^x}{(x-1)^2}+c

 

Posted by

Ravindra Pindel

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