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.

$$\frac{1}{(1+x^2)^2} =  \left(\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}\right)= \left(\frac{1}{1+x^2}+\frac x2\left(\frac{1}{ 1+x^2 }\right)'\right)$$$$=\left(\frac{1}{1+x^2}+\left(\frac 12 \frac {x}{1+x^2}\right)'-\frac 12 \frac {1}{1+x^2}\right) = \frac 12 \left(\arctan x+\frac{x}{1+x^2}\right)'.$$
