Answer:
Option D
Explanation:
Let I= $\int_{0}^{1/2} \frac{1+\sqrt{3}}{((x+1)^{2}(1-x)^{6})^{1/4}}dx$
$\Rightarrow$ I = $\int_{0}^{1/2} \frac{1+\sqrt{3}}{[(1-x)^{2}[(\frac{1-x}{1+x})^{6}]^{1/4}}dx$
Put $\frac{I-x}{I+x}=t \Rightarrow \frac{-2dx}{(1+x)^{2}}\Rightarrow dt$
when x=0 , t=1, x= $\frac{1}{2}$ , t= $\frac{1}{3}$
$\therefore$ I= $\int_{0}^{1/3} \frac{\left(1+\sqrt{3}\right)dt}{-2(t)^{\frac{6}{4}}}$
$\Rightarrow$ $I= \frac{-(1+\sqrt{3})}{2}\begin{bmatrix} \underline{-2} \\ \sqrt{t} \end{bmatrix}^{1/3}_{1}$
$\Rightarrow$ $I= (1+\sqrt{3}) (\sqrt{3}-1)\Rightarrow I=3-1=2$
