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? 

The general formula (for $m > n+1$ and $n \ge 0$) is $\frac{\pi}{m} \csc\left(\frac{\pi (n+1)}{m}\right)$.  IIRC the usual method involves a wedge-shaped contour of angle $2 \pi/m$.
EDIT: Consider $\oint_\Gamma f(z)\ dz$ where $f(z) = \frac{z^n}{1+z^m}$ (using the principal branch if $m$ or $n$ is a non-integer) and $\Gamma$ is the closed contour below:

$\Gamma_1$ goes to the right along the real axis from $\epsilon$ to $R$, so $\int_{\Gamma_1} f(z)\ dz = \int_\epsilon^R \frac{x^n\ dx}{1+x^m}$.  $\Gamma_3$ comes in along the ray at angle $2 \pi/m$.  Since $e^{(2 \pi i/m) m} = 1$, $\int_{\Gamma_3} f(z)\ dz = - e^{2 \pi i (n+1)/m} \int_{\Gamma_1} f(z)\ dz$.  $\Gamma_2$ is a circular arc at distance $R$ from the origin.  Since $m > n+1$, the integral over it goes to $0$ as $R \to \infty$. Similarly, the integral over the small circular arc at distance $\epsilon$ goes to $0$ as $\epsilon \to 0$.  So we get
$$   \lim_{R \to \infty, \epsilon \to 0} 
\int_\Gamma f(z)\ dz =  (1 - e^{2 \pi i (n+1)/m}) \int_0^\infty \frac{x^n\ dx}{1+x^m}$$
The meromorphic function $f(z)$ has one singularity inside $\Gamma$, a pole at $z = e^{\pi i/m}$ where the residue is $- e^{\pi i (n+1)/m}/m$.  So the residue theorem gives you
$$ \int_0^\infty \frac{x^n\ dx}{1+x^m} = \frac{- 2 \pi i e^{\pi i (n+1)/m}}{ m (1 - e^{2 \pi i (n+1)/m})} = \frac{\pi}{m} \csc\left(\frac{\pi(n+1)}{m}\right)$$
