How to solve following integral I am trying to solve following integral.
$$=\int_0^\infty\frac{x}{x^n+(\epsilon+\sigma)}dx$$
I started by assuming that  $a = \epsilon + \sigma$ and $$\implies x^n = A\tan^2(\theta)$$
Then,
$$\implies x = a^{1/n}\tan^{2/n}(\theta)$$
$$\implies dx = \frac{2a^{1/n}}{n} (\tan(\theta))^{2/n-1}\sec^2(\theta)\text{ d}\theta$$
Using $\tan^2(\theta)+1 = \sec^2(\theta)$ and substituing above equations, we can write Eq. 1 as:
$$=\frac{\frac{2a^{2/n}}{n}(\tan(\theta))^{4/n-1}\sec^2(\theta)\text{ d}\theta}{\sec^2(\theta)}$$
$$=\int_0^{\pi/2}\frac{2a^{2/n}}{n}(\tan(\theta))^{4/n-1} \text{ d}\theta$$
let $B = \frac{2a^{2/n}}{n}$, then
$$=\int_0^{\pi/2}B(\tan(\theta))^{4/n-1} \text{ d}\theta$$
Now, I don't know how to proceed forward

The Final answer by authors is:
$\large \frac{\pi\sigma(\epsilon+\sigma)^{2/n-1}}{n\sin(\frac{2\pi}{n})}$
 A: Hint :
Notice that on substitution $x=(at)^{1/n}$ where $a=\epsilon +\sigma$ the integral changes to $$I= \frac {a^{\frac 2n -1}}{n} \int_0^{\infty} \frac {t^{\frac 2n -1}}{1+t}dt $$ 
Notice that the integral part is just $B\left( \frac 2n,  1-\frac 2n\right)$ 
Where $B(x,y)$ is standard Beta Function.  
Using relation between Gamma function and Beta function and then using the Euler's Reflection formula for the Gamma function i.e. $$\Gamma(z)\Gamma(1-z)=\frac {\pi}{\sin (\pi z)}$$ you could evaluate the integral. 
A: Observe we have
\begin{align}
\int^\infty_0 \frac{x}{x^n+a}\ dx = a^{(2-n)/n}\int^\infty_0 \frac{y}{1+y^n}\ dy.
\end{align}
Hence it suffices to evaluate
\begin{align}
\int^\infty_0 \frac{y}{1+y^n}\ dy.
\end{align}
The easiest way is to use contour integration. Observe
\begin{align}
\int^R_0 \frac{y}{1+y^n}\ dy + \int_{C_R} \frac{z}{1+z^n}\ dz - \int_0^R \frac{(R-t)e^{i4\pi/n}}{1+(R-t)^n}dt = 2\pi i \operatorname{Res}\left(\frac{z}{1+z^n}, z_0=e^{i\pi/n} \right)
\end{align}
where $C_R$ is the arc of the circle of radius $R$ and angle $2\pi/n$. Hence we see that
\begin{align}
(1-e^{i4\pi/n}) \int^R_0 \frac{x}{1+x^n}\ dx +\int_{C_R} \frac{z}{1+z^n}\ dz= 2\pi i\frac{z_0}{nz_0^{n-1}} = \frac{2\pi i}{ne^{i\pi (n-2)/n}}
\end{align}
which means
\begin{align}
\int^\infty_0 \frac{x}{1+x^n}\ dx =& \frac{2\pi i}{ne^{i\pi(n-2)/n}(1-e^{4i\pi/n})}= \frac{2\pi i}{ne^{i\pi }(e^{-2\pi i/n}-e^{2\pi i/n})} \\
=& \frac{\pi}{n\sin\left(\frac{2\pi}{n} \right)}.
\end{align} 
Finally, combining everything yields
\begin{align}
\int^\infty_0 \frac{x}{x^n+(\sigma+\varepsilon)}\ dx = \frac{\pi (\sigma+\varepsilon)^{2/n-1}}{n\sin\left(\frac{2\pi}{n} \right)}.
\end{align}
