Question

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

Answer 1

${\int }_0^{\frac{1}{2}}\frac{(1+\sqrt{3})dx}{\left[\right(1+x{)}^2(1-x{)}^6{]}^{1/4}}$

$\Rightarrow {\int }_0^{\frac{1}{2}}\frac{(1+\sqrt{3})dx}{(1+x{)}^{\frac{1}{2}}[(1-x{)}^6{]}^{1/4}}$

$\Rightarrow {\int }_0^{\frac{1}{2}}\frac{(1+\sqrt{3})dx}{(1+x{)}^2{\left[\frac{(1-x{)}^6}{(1+x{)}^6}\right]}^{1/4}}$

Put $\frac{1-x}{1+x}=t\Rightarrow \frac{-2dx}{(1+x{)}^2}=dt$

$I={\int }_1^{1/3}\frac{(1+\sqrt{3})dt}{-2t^{6/4}}=\frac{-(1+\sqrt{3})}{2}\times {\left|\frac{-2}{\sqrt{t}}\right|}_1^{1/3}=(1+\sqrt{3})(\sqrt{3}-1)=2$.