How to integrate $\int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx$

Question

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 $0$

Answer 1

Hint Try substituting $x^{\frac{k}{2}}=\tan\theta$ and watch the integral fall apart.

Answer 2

Setting $$ z=x^{\frac{k}{2}}, $$ we have $$ \int\frac{x^{\frac{k}{2}-1}}{1+x^k}\,dx=\frac2k\int\frac{1}{1+z^2}\,dz=\frac2k\arctan z+c=\frac2k\arctan(x^{\frac{k}{2}})+c, $$ with $c$ an arbitrary constant.

Answer 3

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}