$\int_0^\infty {x^a\over (x^2+1)^2} dx$ where $0$\int_0^\infty {x^a\over (x^2+1)^2} dx$ where $0<a<1$.
I know I can use partial fraction decomposition to obtain two different integrals, but I'm not sure how to integrate them. Any solutions or hints are greatly appreciated.

If you know about the $\text{B}$ and $\Gamma$ functions, you can first let $x^2=u$ and then $t=\frac1{1+u}$ to get
$$\begin{align}\int_9^{\infty}\frac{x^a}{(x^2+1)^2}dx&=
\frac12\int_0^{\infty}\frac{u^{\frac{a-1}2}}{(1+u)^2}du
=\frac12\int_0^1(1-t)^{\frac{a-1}2}t^{\frac{1-a}2}dt\\
&=\frac12\text{B}\left(\frac{a+1}2,\frac{3-a}2\right)
=\frac12\frac{\Gamma\left(\frac{a+1}2\right)\Gamma\left(\frac{3-a}2\right)}{\Gamma\left(2\right)}\\
&=\frac12\Gamma\left(\frac{a+1}2\right)\left(\frac{1-a}2\right)\Gamma\left(\frac{1-a}2\right)\\
&=\frac{1-a}4\cdot\frac{\pi}{\sin\left(\frac{\pi(1-a)}2\right)}
=\frac{\pi(1-a)}{4\cos\left(\frac{\pi a}2\right)}\end{align}$$
