MCQ
$\int_0^1 \sqrt{\frac{1+x}{1-x}} d x=$
  • $\frac{\pi}{2}+1$
  • B
    $\frac{1}{2}$
  • C
    $\frac{\pi}{2}-1$
  • D
    $\frac{\pi}{2}$

Answer

Correct option: A.
$\frac{\pi}{2}+1$
(A)
$\int_0^1 \sqrt{\frac{1+x}{1-x}} d x=\int_0^1 \sqrt{\frac{(1+x)^2}{(1+x)(1-x)}} d x$
$\begin{array}{l}=\int_0^1 \frac{1+x}{\sqrt{1-x^2}} d x \\ =\int_0^1 \frac{1}{\sqrt{1-x^2}} d x+\frac{1}{2} \int_0^1 \frac{2 x}{\sqrt{1-x^2}} d x \\ =\left[\sin ^{-1} x\right]_0^1-\left[\sqrt{1-x^2}\right]_0^1=\frac{\pi}{2}+1\end{array}$

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