Discussion Forum

The value of integral $\int_{0}^{1/2} \frac{1+\sqrt{3}}{((x+1)^{2}(1-x)^{6})^{1/4}}dx$ is ............

Answer:

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= $\int_{0}^{1/3} \frac{\left(1+\sqrt{3}\right)dt}{-2(t)^{\frac{6}{4}}}$

  $\Rightarrow$  $I= \frac{-(1+\sqrt{3})}{2}\begin{bmatrix}  \underline{-2} \\  \sqrt{t} \end{bmatrix}^{1/3}_{1}$

$\Rightarrow$       $I= (1+\sqrt{3}) (\sqrt{3}-1)\Rightarrow I=3-1=2$