Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$ I've been looking at
$$\int\limits_0^\infty  {\frac{{{x^n}}}{{1 + {x^m}}}dx }$$
It seems that it always evaluates in terms of $\sin X$ and $\pi$, where $X$ is to be determined. For example:
$$\displaystyle \int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}$$
$$\int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}$$
$$\int\limits_0^\infty  {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}$$
So I guess there must be a closed form - the use of $\Gamma(x)\Gamma(1-x)$ first comess to my mind because of the $\dfrac{{\pi x}}{{\sin \pi x}}$ appearing. Note that the arguments are always the ratio of the exponents, like $\dfrac{1}{4}$, $\dfrac{1}{3}$ and $\dfrac{2}{5}$. Is there any way of finding it? I'll work on it and update with any ideas.

UPDATE:
The integral reduces to finding 
$$\int\limits_{ - \infty }^\infty  {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt} $$
With $a =\dfrac{n+1}{m}$ which converges only if 
$$0 < a < 1$$
Using series I find the solution is

$$\sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} $$

Can this be put it terms of the Digamma Function or something of the sort? 

Contour Integration Approach
Assuming only that $m>0$ and $-1<n<m-1$, let $z=x^m$ with the exaggerated contour $\gamma$:

$\gamma$ is actually tight above and below the positive real axis and around the origin and circles back at an arbitrarily large distance from the origin.
The part just above the positive real axis captures the integral. The part just below the positive real axis gets $-e^{2\pi i\frac{n-m+1}{m}}$ times the integral. The residue at $z=-1$ of $\frac{z^{\frac{n-m+1}{m}}}{1+z}$ is $e^{\pi i\frac{n-m+1}{m}}$. Putting this all together yields
$$
\frac1m\int_\gamma\frac{z^{\frac{n-m+1}{m}}}{1+z}\,\mathrm{d}z=\left(1-e^{2\pi i\frac{n-m+1}{m}}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
$$
\frac{2\pi i}{m}e^{\pi i\frac{n-m+1}{m}}=\left(1-e^{2\pi i\frac{n-m+1}{m}}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
$$
\frac{\pi}{m}=\sin\left(\pi\frac{n+1}{m}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
$$
\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x\tag1
$$

Relation to $\bf{\Gamma(\alpha)\Gamma(1-\alpha)}$
Setting $m=1$, $n=\alpha-1$, and using the substitution $s=tu$ yields Euler's Reflection Formula:
$$
\begin{align}
\Gamma(\alpha)\Gamma(1-\alpha)
&=\int_0^\infty s^{\alpha-1}e^{-s}\,\mathrm{d}s\int_0^\infty t^{-\alpha}e^{-t}\,\mathrm{d}t\\
&=\int_0^\infty\int_0^\infty u^{\alpha-1}e^{-(tu+t)}\,\mathrm{d}u\,\mathrm{d}t\\
&=\int_0^\infty\frac{u^{\alpha-1}}{1+u}\,\mathrm{d}u\\[6pt]
&=\pi\csc(\pi\alpha)\tag2
\end{align}
$$

Another Proof of $\bf{(1)}$
$$
\begin{align}
\int_0^\infty\frac{x^{\alpha-1}}{1+x}\,\mathrm{d}x
&=\int_0^1\frac{x^{-\alpha}+x^{\alpha-1}}{1+x}\,\mathrm{d}x\\
&=\sum_{k=0}^\infty(-1)^k\int_0^1\left(x^{k-\alpha}+x^{k+\alpha-1}\right)\mathrm{d}x\\
&=\sum_{k=0}^\infty(-1)^k\left(\frac1{k-\alpha+1}+\frac1{k+\alpha}\right)\\
&=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+\alpha}\\[9pt]
&=\pi\csc(\pi\alpha)\tag3
\end{align}
$$
where the last step of $(3)$ is proven in $(3)$ of this answer. Then apply $(3)$ to get
$$
\begin{align}
\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
&=\frac1m\int_0^\infty\frac{x^{\frac{n-m+1}m}}{1+x}\,\mathrm{d}x\\
&=\frac\pi{m}\csc\left(\pi\frac{n+1}m\right)\tag4
\end{align}
$$
