Question

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

Answer 1

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

Put $x+1=u$ and $x^2=(u-1{)}^2$

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

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

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

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

$I=e^{u-1}+2e^{-1}\int \frac{1}{u^2}{e}^{u}du–2^{-1}\frac{1}{u}{e}^u+2e^{-1}\int -\frac{1}{u^2}{e}^{u}du$

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

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

$I=\frac{e^x(x+1)–2e^x}{x+1}+c$

$I=\frac{e^{x}x-e^x}{x+1}+c=\frac{e^x(x-1)}{x+1}+c$