Question:
Evaluate the following integrals:
$\int e^{x} \frac{(x-4)}{(x-2)^{3}} d x$
Solution:
Let $I=\int e^{x} \frac{(x-4)}{(x-2)^{3}} d x$
$=\int e^{x} \frac{(x-2)-2)}{(x-2)^{3}} d x$
$=\int e^{x}\left\{\frac{1}{(x-2)^{2}}-\frac{2}{(x-2)^{2}}\right\} d x$
Let $f(x)=\frac{1}{(x-2)^{2}}$ and $f^{\prime}(x)=\frac{2}{(x-2)^{2}}$
We know that, $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\}=e^{x} f(x)+c$
$I=\frac{e^{x}}{(x-2)^{2}}+c$
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