Evaluate ∫1(x2+1)2dx

${\displaystyle \int \frac{dx}{{(x^2+1)}^2}}$
Set ${\displaystyle x=\tan \left(u\right)}$ and ${\displaystyle dx={\sec }^2\left(u\right)du}$. Then ${\displaystyle {(x^2+1)}^2={({\tan }^2\left(u\right)+1)}^2={\sec }^4\left(u\right)}$ and ${\displaystyle u=\arctan \left(x\right)}$
${\displaystyle =\int {\cos }^2\left(u\right)du=\frac{1}{2}\int \cos \left(2u\right)du+\frac{1}{2}\int 1du}$
${\displaystyle =\frac{u}{2}+\frac{1}{4}\sin \left(2u\right)+C}$
${\displaystyle =\frac{x^2\arctan \left(x\right)+x+\arctan \left(x\right)}{2x^2+2}+C}$

There is a faster way. Substitute
${\displaystyle x=\tan \left(z\right)dx={\sec }^2\left(z\right)dz}$
thus
${\displaystyle {(x^2+1)}^2={({\tan }^2\left(z\right)+1)}^2={\sec }^4\left(z\right)z=\arctan \left(x\right)}$
And remembering that
${\displaystyle \frac{1}{{\sec }^2}={\cos }^2}$
your integral is simply
${\displaystyle \int {\cos }^2\left(z\right)dz}$
Which is trivial and left to you.
The tangent/secant substitution is a great technique which most of people ignore. Study it, and you will solve lots of awesome integrals!
Final result:
${\displaystyle \frac{x^2\arctan \left(x\right)+x+\arctan \left(x\right)}{2x^2+2}}$

Let us find $\frac{d[(ax+b)/(x^2+1{)}^n]}{dx}$
$$\begin{gathered} =\frac{-a{x}^2-2bx+a}{(x^2+1{)}^{n+1}} \\ =\frac{-a(x^2+1)-2bx+2a}{(x^2+1{)}^{n+1}} \\ =-\frac{a}{(x^2+1{)}^n}+\frac{2a-2bx}{(x^2+1{)}^{n+1}} \end{gathered}$$
Integrating both sides,
$$\begin{gathered} b=0\Rightarrow 2\int \frac{dx}{(x^2+1{)}^{n+1}}=\int \frac{dx}{(x^2+1{)}^n}+\frac{x}{(x^2+1{)}^n} \\ n=1\Rightarrow 2\int \frac{dx}{(x^2+1{)}^2}=\int \frac{dx}{x^2+1}+\frac{x}{(x^2+1)} \end{gathered}$$