Question

Evaluate

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

Answer 1

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

Substitute $u=x+1$ $\to dx=du$ & $x^2=(u-1{)}^2$

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

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

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

$=e^{-1}\left[\int \frac{2e^u}{u}dy+\int \frac{e^u}{u^2}du+\int e^{u}du+\frac{e^u}{u^2}du\right]$

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

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

$=e^{-1}[-Ei(u)-\frac{e^u}{u}+3^u+Ei(u)-\frac{e^u}{u}]$

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

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

$=e^x-\frac{2e^x}{x+1}+c$.