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.

Hint:
On this page, you find the reduction formula
$$\int \frac{dx}{(x^2 +m^2)^k} = x\frac1{2 m^2 (k-1) (x^2 +m^2)^{k-1}}+
\frac{2k-3}{2m^2 (k-1)} \int \frac{dx}{(x^2 + m^2)^{k-1}}$$
which you can use for your integral...
