Integral of $\frac{1}{(1+x^2)^2}$ I am in the middle of a problem and having trouble integrating the following integral:
$$\int_{-1}^1\frac1{(1+x^2)^2}\mathrm dx$$
I tried doing partial fractions and got:
$$1=A(1+x^2)+B(1+x^2)$$
I have no clue how to solve this since it is obvious there is no way to cancel out either $A$ or $B$ to get the other variable. Please guide me.
Thank you.

$$
\begin{aligned}
\int \frac{1}{\left(1+x^{2}\right)^{2}} d x\stackrel{IBP}{=} &-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^{2}}\right) \\=&-\frac{1}{2 x\left(1+x^{2}\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^{2}}-\frac{1}{x^{2}}\right) d x \\=&-\frac{1}{2 x\left(1+x^{2}\right)}+\frac{1}{2} \tan ^{-1} x+\frac{1}{2 x}+C\\
=&\frac{1}{2}\left(\frac{x}{1+x^{2}}+\tan ^{-1} x\right)+C
\end{aligned}
$$
