Question

The value of the integral ${\int }_0^{\frac{1}{2}}\frac{1+\sqrt{3}}{{\left({\left(x+1\right)}^2{\left(1-x\right)}^6\right)}^{\frac{1}{4}}}dx$ is

Answer 1

Answer: 2

${\int }_0^{\frac{1}{2}}\frac{\left(1+\sqrt{3}\right)dx}{{\left(\left[{\left(1+x\right)}^2{\left(1-x\right)}^6\right]\right)}^{\frac{1}{4}}}$
${\int }_0^{\frac{1}{2}}\frac{\left(1+\sqrt{3}\right)dx}{{\left(1+x\right)}^2{\left(\left[\frac{{\left(1-x\right)}^6}{{\left(1+x\right)}^6}\right]\right)}^{\frac{1}{4}}}$
Put $\frac{1-x}{1+x}=t\Rightarrow \frac{-2dx}{{\left(1+x\right)}^2}=dt$
$I={\int }_1^{\frac{1}{3}}\frac{\left(1+\sqrt{3}\right)dt}{-2t^{\frac{6}{4}}}=\frac{-\left(1+\sqrt{3}\right)}{2}\times {\left(\left|-\frac{2}{\sqrt{t}}\right|\right)}_1^{\frac{1}{3}}=\left(1+\sqrt{3}\right)\left(\sqrt{3}-1\right)=2$