I am thinking on integrals of the form
$$ \int\limits_0^{\infty} \dfrac{ x^a }{(x+1)(x^2+1) } dx $$
First of all, we have to realize that integral only make sense if $a>2$.
If $a=0$, we can just apply partial fractions, so it is easy.
If $a=1$, then pretty much as before and same with $a=2$.
My question is, how about if $a=1/2$ or $a=3/2$, does the integral have closed form?
Note $\frac{ 1}{(x+1)(x^2+1) } = \frac12\left(\frac{1}{x+1 }+\frac{1-x}{x^2+1 }\right)$ and
$$ \int\limits_0^{\infty} \dfrac{ x^a }{(x+1)(x^2+1) } dx =\frac12 \int\limits_0^{\infty}\left( \dfrac{ x^a }{x+1 } + \frac{ x^{a}}{x^2+1 } -\frac{ x^{a +1}}{x^2+1 } \right)dx\\ =\frac\pi4\left(-2\csc\pi\alpha + \sec\frac{\pi\alpha}2+ \csc\frac{\pi\alpha}2\right) $$
where the result $\int_0^\infty \frac{x^{b-1}}{x^m +1}= \frac\pi m \csc\frac{\pi b}m$ is used.
Using partial fractions, $$\int\limits_0^{\infty} \dfrac{ x^a }{(x+1)(x^2+1) } dx = \frac 1 2 \int\limits_0^{\infty} \dfrac{ x^a }{(x+1) } dx + \frac 1 2 \int\limits_0^{\infty} \dfrac{ x^a }{(x^2+1) } - \frac 1 2 \int\limits_0^{\infty} \dfrac{ x^{a+1} }{(x^2+1) }dx$$ Now using this answer $$\int\limits_{0}^{\infty} \frac{x^{\alpha-1}}{1+x^{\beta}} \ \text{dx} = \frac{\pi}{\beta \sin\left(\pi\frac{\alpha}\beta\right)}, \quad 0 < \alpha < \beta$$ Thus $$\boxed{\int\limits_0^{\infty} \dfrac{ x^a }{(x+1)(x^2+1) } dx = \frac 1 2 \left(\frac {\pi}{\sin(\pi(a+1))} +\frac {\pi}{2\sin\left(\pi\frac{a+1}2\right)} - \frac {\pi}{2\sin\left(\pi\frac{a+2}2\right)}\right)}$$
I shall not repeat the steps already given in answers but give the most compact form I was able to find $$\int\limits_0^{\infty} \dfrac{ x^a }{(x+1)(x^2+1) } dx=\frac{\pi }{2 \left[1+\sqrt{2} \sin \left((2 a+1)\frac{\pi}{4} \right)\right]}$$
