Integrate the function ( x - 3 ) e^x ( x - 1 ) ^3

Question

Integrate the function $\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}$

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Answer

Let $I = \frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}dx$
$= \int {\frac{{\left( {x - 1} \right) - 2}}{{{{\left( {x - 1} \right)}^3}}}{e^x}} dx$
$= \int {{e^x}\left[ {\frac{{\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^3}}} - \frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx$
$\Rightarrow I = \int {{e^x}\left[ {\frac{1}{{{{\left( {x - 1} \right)}^2}}} + \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx$
$\left[ {\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} } \right]$
It is in the form of ${\int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx} }$ since here $f\left( x \right) = \frac{1}{{{{\left( {x - 1} \right)}^2}}}$ and $f'\left( x \right) = \frac{d}{{dx}}\left\{ {{{\left( {x - 1} \right)}^{ - 2}}} \right\}$
$ = - 2{\left( {x - 1} \right)^{ - 3}}$
$ = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}$.
$ \Rightarrow I = \frac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + c$
$\left[ {\because \int {{e^x}\left\{ {f\left( x \right) + f'\left( x \right)} \right\}dx = {e^x}f\left( x \right) + c} } \right]$

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