Question

Evaluate: $\int \frac{(x-4)e^x}{(x-2{)}^3}dx$.

Answer 1

Given the integral

$\int \frac{(x-4)e^x}{{(x-2)}^3}dx=\int \frac{(x-2-2)e^{(x-2)}{e}^2}{{(x-2)}^3}dx=e^2\int \frac{(x-2-2)e^{(x-2)}}{{(x-2)}^3}dx$

Let us assume,

$u=x-2\Rightarrow \frac{du}{dx}=1\Rightarrow du=dx$

Substituting these values in the integral we get,

$e^2\int \frac{(x-2-2)e^{(x-2)}}{{(x-2)}^3}dx=e^2\int \frac{(u-2)e^u}{u^3}du$

For $\int \frac{(u-2)e^u}{u^3}du$,

$\int \frac{(u-2)e^u}{u^3}du=\int \frac{u{e}^u}{u^3}du-2\int \frac{e^u}{u^3}du=\int \frac{e^u}{u^2}du-2\int \frac{e^u}{u^3}du$

Using integration by parts for $\int \frac{e^u}{u^3}du$

$=\frac{2e^u}{2u^2}+2\int -\frac{e^u}{2u^2}du+\int \frac{e^u}{u^2}du=\frac{e^u}{u^2}-\int \frac{e^u}{u^2}du+\int \frac{e^u}{u^2}du=\frac{e^u}{u^2}$

So,

$e^2\int \frac{(u-2)e^u}{u^3}du=e^2\frac{e^u}{u^2}=\frac{e^{u+2}}{u^2}[\because u=x-2]=\frac{e^x}{{(x-2)}^2}\therefore \int \frac{(x-4)e^x}{{(x-2)}^3}dx=\frac{e^x}{{(x-2)}^2}+C.$