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$\int \sqrt{\frac{2+x}{2-x}} d x$ is equal to
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2321 Upvotes
Verified Answer
The correct answer is:
$2 \sin ^{-1}\left(\frac{x}{2}\right)-\sqrt{4-x^2}+C$
Let
$$
\begin{aligned}
I & =\int \sqrt{\frac{2+x}{2-x}} d x \\
& =\int \sqrt{\frac{2+x}{2-x}} \times \sqrt{\frac{2+x}{2+x}} d x=\int \frac{2+x}{\sqrt{4-x^2}} d x \\
& =2 \int \frac{1}{\sqrt{(2)^2-x^2}} d x+\int \frac{x}{\sqrt{4-x^2}} d x \\
& =2 \sin ^{-1} \frac{x}{2}-\sqrt{4-x^2}+C
\end{aligned}
$$
$$
\begin{aligned}
I & =\int \sqrt{\frac{2+x}{2-x}} d x \\
& =\int \sqrt{\frac{2+x}{2-x}} \times \sqrt{\frac{2+x}{2+x}} d x=\int \frac{2+x}{\sqrt{4-x^2}} d x \\
& =2 \int \frac{1}{\sqrt{(2)^2-x^2}} d x+\int \frac{x}{\sqrt{4-x^2}} d x \\
& =2 \sin ^{-1} \frac{x}{2}-\sqrt{4-x^2}+C
\end{aligned}
$$
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