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

The integral expression is given as,

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

Consider

$\frac{1}{(x+1{)}^2\left(x^2+1\right)}=\frac{A}{(x+1{)}^2}+\frac{B}{\left(x^2+1\right)}+\frac{Cx+D}{x^2+1}$

$\Rightarrow 1=A\left(x^2+1\right)+B(x+1)\left(x^2+1\right)+(Cx+D)(x+1{)}^2$

Substitute $x=-1$,

$A=\frac{1}{2}$

Substitute $x^2=-1$.

$1=(Cx+D)\left(x^2+2x+1\right)$

$\Rightarrow 1=(Cx+D)2x$

$\Rightarrow 1=-2C+2Dx$

Substitute $x=0$,

$-\frac{1}{2}=C, D=0$

Comparing coefficient of $x^3$,

$0=B+C$

$\Rightarrow B=\frac{1}{2}$

Now,

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

$=\frac{1}{2}\left[-\frac{1}{x+1}+{\log }_e(x+1)-\frac{1}{2}{\log }_e\left(x^2+1\right)\right]+C$

$={\log }_e\sqrt{x+1}-\frac{1}{2}{\log }_e\sqrt{x^2+1}-\frac{1}{2(x+1)}+C$