JEE Integral

The substitution $x = \cos\theta$ gives $\begin{aligned} \int_0^{\frac12} \frac{1 + \sqrt{3}}{(1+x)^{\frac12}(1-x)^{\frac32}}\,dx & = \; \int_0^{\frac12} \frac{1 + \sqrt{3}}{\sqrt{1-x^2}(1-x)}\,dx \; = \; \int_{\frac12\pi}^{\frac13\pi} \frac{1 + \sqrt{3}}{\sin \theta (1 - \cos\theta)}(-\sin\theta)\,d\theta \\ & = \; \tfrac{1 + \sqrt{3}}{2}\int_{\frac13\pi}^{\frac12\pi} \mathrm{cosec}^2\tfrac12\theta\,d\theta \; = \; (1 + \sqrt{3})\Big[-\cot\tfrac12\theta \Big]_{\frac13\pi}^{\frac12\pi} \\ & = \; (1 + \sqrt{3})(\sqrt{3} - 1) \; = \; \boxed{2} \end{aligned}$

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

$\left.-\frac{\left(\sqrt{3}+1\right) (x-1) (x+1)}{\sqrt[4]{(x-1)^6 (x+1)^2}}\right|_{x=0}^{\frac12}\Rightarrow 2$