Question

The value of the integral $\displaystyle \int _{0}^{\frac{1}{2}} \frac{1 + \sqrt{3}}{\left(\left(x + 1\right)^{2} \left(1 - x\right)^{6}\right)^{\frac{1}{4}}} \, dx$ is

Answer 1

$\displaystyle \int _{0}^{\frac{1}{2}} \frac{\left(1 + \sqrt{3}\right) d x}{\left(\left[\left(1 + x\right)^{2} \left(1 - x\right)^{6}\right]\right)^{\frac{1}{4}}}$
$\displaystyle \int _{0}^{\frac{1}{2}}\frac{\left(1 + \sqrt{3}\right) d x}{\left(1 + x\right)^{2} \left(\left[\frac{\left(1 - x\right)^{6}}{\left(1 + x\right)^{6}}\right]\right)^{\frac{1}{4}}}$
Put $\frac{1 - x}{1 + x}=t\Rightarrow \frac{- 2 d x}{\left(1 + x\right)^{2}}=dt$
$I=\displaystyle \int _{1}^{\frac{1}{3}}\frac{\left(1 + \sqrt{3}\right) d t}{- 2 t^{\frac{6}{4}}}=\frac{- \left(1 + \sqrt{3}\right)}{2}\times \left(\left|- \frac{2}{\sqrt{t}}\right|\right)_{1}^{\frac{1}{3}}=\left(1 + \sqrt{3} \, \right)\left(\sqrt{3} - 1\right)=2$