Can you solve $\int 1/(x^2+1)^2\, dx$ I know that the integral looks like like the anti-derivative of $\arctan$, but i don't know how to use this fact so I tried to use a fraction decomposition to: $$\frac{1}{(x^2+1)^2}$$
to solve the anti-derivative : $$\int \frac{1}{(x^2+1)^2}\, dx$$
so :$$\frac{1}{(x^2+1)^2}=\frac{A}{x^2+1}+\frac{B}{x^2+1}\iff \frac{1}{x^2+1}=A+B$$
$$1+0x^2=(A+B)x^2+(A+B)1$$
Thus I got this system of equations:
$$\cases{A+B=1 \\ A+B=0}$$ Which is impossible. I also tried to use substitution but it didn't help me.

There's multiple ways to do this integral including IBP, substitution... Here's one that uses Feynman's trick (differentiation under the integral sign)
We know the following antiderivative
$$I(a)=\int^x\frac{ dx'}{x'^2+a^2}=\frac{1}{a}\arctan(x/a)$$
You can see we introduced a parameter that did not exist before. We differentiate under the integral sign to obtain the desired integral form
$$J(a)=\int^x\frac{dx'}{(x'^2+a^2)^2}=-\frac{1}{2a}\frac{\partial I}{\partial a}=\frac{1}{2a^3}\left(\arctan(x/a)+\frac{a x}{x^2+a^2}\right)$$
$J(1)$ is the indefinite integral we seek.
EDIT: Here's a way that utilizes IBP.
Rewrite by adding and subtracting $x^2$ in the numerator
$$\int\frac{dx}{(x^2+1)^2}=\int \frac{dx}{x^2+1}-\int\frac{x^2dx}{(x^2+1)^2}$$
However note that
$$\int xdx\frac{x}{(x^2+1)^2}=-\frac{1}{2}\int dx~x \frac{d}{dx}\left(\frac{1}{x^2+1}\right)=-\frac{x}{2(x^2+1)}+\frac{1}{2}\arctan x$$
which yields the desired result
$$\int\frac{dx}{(x^2+1)^2}=\frac{1}{2}\arctan x+\frac{1}{2}\frac{x}{x^2+1}$$
