JEE Advanced 2018 · Paper 2, Mathematics question 7

Simplify. $\left((x+1)^2(1-x)^6\right)^{1/4}=(1+x)^{1/2}(1-x)^{3/2}=(1-x)^2\sqrt{u}$, where $u=\dfrac{1+x}{1-x}$.

Substitute. $du=\dfrac{2\,dx}{(1-x)^2}$, and $u$ runs from 1 to 3:
$$\int_0^{1/2}\frac{dx}{(1-x)^2\sqrt u}=\int_1^3\frac{du}{2\sqrt u}=\sqrt3-1.$$

Value.
$$(1+\sqrt3)(\sqrt3-1)=2.$$

Answer: 2