Calculate using Euler integrals $\int \limits_{-1}^{1} \frac{dx}{\sqrt[3]{(1+x)^2(1-x)}}$

Question

Calculate using Euler integrals $$\int \limits_{-1}^{1} \frac{dx}{\sqrt[3]{(1+x)^2(1-x)}}$$

What substitution is better to use here?

I have tried something like $t = \frac{1-x}{1+x}$ but it doesn't look good.

Answer 1

If you set $x=2t-1$ you get

$$ 2\int_{0}^{1}\frac{dt}{\sqrt[3]{(2t)^2(2-2t)}}=\int_{0}^{1} t^{-2/3}(1-t)^{-1/3}\,dt = \frac{\Gamma(1/3)\,\Gamma(2/3)}{\Gamma(1)} = \frac{\pi}{\sin\frac{\pi}{3}}=\color{red}{\frac{2\pi}{\sqrt{3}}}.$$

Answer 2

In general, $$\int_{-1}^1 \frac{d x}{\sqrt[n]{(1+x)^{n-1}(1-x)}} = \pi \csc \left(\frac{\pi}{n}\right) $$ Proof:

Letting $x=\cos 2 \theta$, then $d x=-2 \sin 2 \theta d \theta$ $=-4 \sin \theta \cos \theta d \theta$ and $$ \begin{aligned} I & =\int_{\frac{\pi}{2}}^0 \frac{-4 \sin \theta \cos \theta d \theta}{\sqrt[n]{\left(2 \cos ^2 \theta\right)^{n-1}\left(2 \sin ^2 \theta\right)}} \\ & =2 \int_0^{\frac{\pi}{2}} \cos ^{2\left(\frac{1}{n}\right)-1} \theta \sin ^{2\left(1-\frac{1}{n}\right)-1} \theta d \theta \\ & =B\left(\frac{1}{n},1-\frac{1}{n}\right) \\ & =\pi \csc \left(\frac{\pi}{n}\right) \quad \text { By Reflection Property of Beta function} \end{aligned} $$

In particular, $$ I_3=\pi \csc \left(\frac{\pi}{3}\right)=\frac{2 \pi}{\sqrt{3}} \textrm{ and } I_6=\pi \csc \left(\frac{\pi}{6}\right)=2 \pi $$

Answer 3

$$\int \limits_{-1}^{1} \frac{dx}{\sqrt[3]{(1+x)^2(1-x)}} \overset{x=\frac{1-t^3}{1+t^3}}= 3\int_0^\infty \frac {t}{1+t^3}dt \overset{t\to\frac{1}{t}} =\frac32\int_0^\infty \frac{1+t}{1+t^3}dt\\ =\frac32\int\frac{dt}{1-t+t^2}=\frac{2\pi}{\sqrt3}$$