If we put $I:=\int_0^\infty\frac{x^p}{1+x^2}dx$ making the following variable change we get $x=-u\Longrightarrow dx=-du\Longrightarrow \int_{-\infty}^0\frac{x^p}{1+x^2}dx=\int_\infty^0\frac{(-1)^pu^p}{1+u^2}(-du)=(-1)^pI$ so that $\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\left(1+(-1)^p\right)I=\left(1+e^{p\pi i}\right)I\Longleftrightarrow$
$\Longleftrightarrow \,\,(***)\,\,\,I=\Re\left(\frac{1}{1+e^{p\pi i}}\right)\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\frac{1}{2}\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx$
For $\,p\geq 0\,$ let us try the following contour, with $\,1: $C:=[-R,R]\cup\left(\gamma_R:=\left\{z\in\Bbb C\;:\;z=Re^{it}\,\,,0\leq t\leq\pi\right\}\right)$
Putting $f(z):=\frac{z^p}{1+z^2}\Longrightarrow Res_{z=i}(f)=\lim_{z\to i}(z-i)f(z)=\frac{i^p}{2i}=\frac{e^{p\pi i/2}}{2i}$ so by CIT we get $\oint_Cf(z)\,dz=2\pi i\frac{1}{2i}e^{p\pi i/2}=\pi \left(\cos\frac{p\pi}{2}+i\sin\frac{p\pi}{2}\right)$
Now, by the estimation lemma, we get $\left|\int_{\gamma_R}f(z)\,dz\right|\leq \max_{z\in\gamma_R}\frac{|z|^p}{|1+z^2|}R\pi\leq \frac{\pi R^{p+1}}{1-R^2}\xrightarrow [R\to\infty]{} 0\,\,,\,\text{since}\,\,p+1<2$
Thus, taking the limit above, we get $(@@@)\,\,\,\pi \left(\cos\frac{p\pi}{2}+i\sin\frac{p\pi}{2}\right)=\lim_{R\to\infty}\oint_Cf(z)dz=\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx\Longrightarrow$ $\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\pi\cos\frac{p\pi}{2}\Longleftrightarrow$
$\stackrel{\text{from (***) above}}\Longleftrightarrow I=\frac{\pi\cos p\frac{\pi}{2}}{2}\,\,\,\,\,Q.E.D.$
Now, for $\,-1< p<0\,$ all the above remains mutatis mutandi but there's also the matter of the pole at $\,z=0\,$ ,which I can't handle as I can't manage to find out what is this pole's multiplicity. Yet I'm almost sure the residue here is zero, but can't prove it.
Added: As did proposes in a comment below, let's use the following change of variables: $x=\frac{1}{u}\Longrightarrow dx=-\frac{du}{u^2}\Longrightarrow I(p):=\int_\infty^0\frac{u^{-p}}{1+\frac{1}{u^2}}\left(-\frac{du}{u^2}\right)=:I(-p)$ so that all we did above indeed remains valid when $-1< p<0\,$
Added trying to address Joriki's point in the comments below: We got, 3 lines before $\,(***)\,$, that $\int_{-\infty}^0\frac{x^p}{1+x^2}dx=e^{ip\pi}I\Longrightarrow \int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\int_{-\infty}^0\frac{x^p}{1+x^2}dx+\int_0^\infty\frac{x^p}{1+x^2}dx=$ $=(1+e^{ip\pi})I$ and as Joriki remarks from this it follows $I=\frac{1}{1+e^{ip\pi}}\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx$ without taking the real part in the RHS , and thus in $\,(@@@)\,$ we actually have $\int_{-\infty}^\infty\frac{x^p}{1+x^2}dx=\pi\left(\cos p\frac{\pi}{2}+i\sin p\frac{\pi}{2}\right)=\pi e^{ip\pi}$ and from here we get the same result as Mercy got below: $I=\frac{\pi e^{ip\pi}}{1+e^{ip\pi}}=\frac{\pi}{2}\sec p\frac{\pi}{2}$