$\int_0^\infty \frac{x^a}{(x^2+1)^2}dx$ The following exercise is from the book of Churchill of complex analysis. Solve the integral
$$\int_0^\infty \frac{x^a}{(x^2+1)^2}dx$$
where $-1<a<3$ the book of complex analysis of Churchill suggests the indented countour: 
And also the book gives you the answer $\frac{\pi(1-a)}{4\cos(a\pi/2)}$.
Progress: I did $\int_\Gamma \frac{z^a}{(z^2+1)}dz$ where $\Gamma$ is the region above. I already have
$$\int_\Gamma \frac{z^a}{(z^2+1)^2}dz=\frac{i\pi e^{ai\pi/2}(1-a)}{2},$$
$$\int_\Gamma \frac{z^a}{(z^2+1)^2}dz=\int_{-R}^{-r} \frac{z^a}{(z^2+1)^2}dz+\int_r^R \frac{z^a}{(z^2+1)^2}dz-\int_{C_r} \frac{z^a}{(z^2+1)^2}dz+\int_{C_R} \frac{z^a}{(z^2+1)^2}dz.$$
$$\lim_{R\rightarrow\infty}\int_{C_R} \frac{z^a}{(z^2+1)^2}dz=0.$$
and
$$\lim_{R\rightarrow\infty,r\rightarrow 0}\left(\int_{-R}^{-r} \frac{z^a}{(z^2+1)^2}dz+\int_r^R \frac{z^a}{(z^2+1)^2}dz\right)=(1+(-1)^a)\int_0^\infty\frac{z^a}{(z^2+1)^2}dz=(1+e^{i\pi a})\int_0^\infty\frac{z^a}{(z^2+1)^2}dz$$
So I get
$$\int_0^\infty\frac{z^a}{(z^2+1)^2}dz=\frac{\pi(1-a)}{4\cos(a\pi/2)}+\lim_{r\rightarrow 0}\int_{C_r} \frac{z^a}{(z^2+1)^2}dz.$$
Where $C_R$ is the upper semicircle of radius $R$ with positive orientation and $C_r$ is the upper semicircle of radius $r$ with positive orientation.
So that tells me it must happen that
$$\lim_{r\rightarrow 0}\int_{C_r} \frac{z^a}{(z^2+1)^2}dz=0,$$
but I haven't been able to prove it, I have tried to use Jordan's lemma doing a change of variable, integrating and using L'Hopital and I don't get that this integral converges to $0$. Does somebody knows how to achieve it?

$$ \left | \int_{C_r}\frac{z^a}{(z^2+1)^2} dz \right |\le \int_{C_r} \left | \frac{z^a}{(z^2+1)^2} dz \right | \le \int_{C_r}\frac{r^a}{(1-r^2)^2} |dz|= \frac{\pi r^{a+1}}{(1-r^2)^2} \to 0$$
as $a+1 \gt 0$.
