Question

Find $\int \frac{(x^2+1)e^x}{(x+1{)}^2}dx$.

Answer 1

Let $I=\int \frac{(x^2+1)e^x}{(x+1{)}^2}dx$

Substitute $u=x+1⟹du=dx,x^2=(u-1{)}^2$

$⟹I=e^{-1}\int \frac{\left(\right(u-1{)}^2+1)e^u}{u^2}du$

$=e^{-1}\int \left(\frac{(u-1{)}^2{e}^u}{u^2}\right)du+\int \frac{e^u}{u^2}du$

$=e^{-1}\int \left(\frac{(u^2-2u+1{)}^2{e}^u}{u^2}\right)du+\int \frac{e^u}{u^2}du$

$=e^{-1}\int (e^u-\frac{2e^u}{u}+\frac{e^u}{u^2})du+\int \frac{e^u}{u^2}du$

$=e^{-1}\int (e^u-\frac{2e^u}{u}+\frac{2e^u}{u^2})du$

$⟹I=e^{-1}(e^u-2E_i(u)+2\int \frac{e^u}{u^2}du)$ ........ $\left(2\right)$

To evaluate $\int \frac{e^u}{u^2}du$

Let $f=e^u$ and $g^{\prime }=\frac{1}{u^2}⟹f^{\prime }=e^u$ and $g=-\frac{1}{u}$

Using integration byparts

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

$=-\frac{e^u}{u}+E_i\left(u\right)$

From $\left(2\right)$,

$I=e^{-1}(e^u-2E_i(u)-2\frac{e^u}{u}+2E_i(u\left)\right)$

$=e^{u-1}-\frac{2e^{u-1}}{u}$

$⟹I=e^x-\frac{2e^x}{x+1}+C$

$⟹I=\frac{e^x(x-1)}{x+1}+C$