How to integrate $\int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx$ How would I do the following integral? 
$$\int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx$$
Where $x > 0$ and $k$ is a constant greater than $0$

Consider the integral
\begin{align}
I = \int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx
\end{align}
Let $t = x^{k/2}$ for which $x = t^{2/k}$ and $dx = (2/k) t^{(2/k) - 1} \, dt$ for which the integral becomes
\begin{align}
I = \frac{2}{k} \int \frac{dt}{1 + t^{2}}.
\end{align} 
This is the integral for $\tan^{-1}(t)$ leading to
\begin{align}
I = \frac{2}{k} \, \tan^{-1}(t) + c_{0}
\end{align}
and upon backward substitution
\begin{align}
\int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx = \frac{2}{k} \, \tan^{-1}\left(x^{\frac{k}{2}}\right) + c_{0}
\end{align}
