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$\int \sqrt{\frac{1+x}{1-x}} d x=$
(where $C$ is a constant of integration.)
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(where $C$ is a constant of integration.)
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1798 Upvotes
Verified Answer
The correct answer is:
$\sin ^{-1} x-\sqrt{1-x^2}+C$
$\begin{aligned} & \int \sqrt{\frac{1+x}{1-x}} \mathrm{~d} x=\int \sqrt{\frac{(1+x)^2}{1-x^2}} \mathrm{~d} x=\int\left\{\frac{1}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}\right\} \mathrm{d} x \\ & =\sin ^{-1} x-\frac{1}{2} \times 2 \sqrt{1-x^2} \\ & =\sin ^{-1} x-\sqrt{1-x^2}+C\end{aligned}$
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