Question

${\int }_0^2\sqrt{\frac{2+x}{2-x}}dx$ is equal to

• A. $\pi +1$
• B. $1+\pi /2$
• C. $\pi +3/2$
• D. none of these

Answer 1

The correct option is D none of these

Let $I=\sqrt{\frac{2+x}{2-x}}dt$
Put $t=\frac{2+x}{2-x}\Rightarrow dt=(\frac{1}{2-x}-\frac{-x-2}{{(2-x)}^2})dx$
$\therefore I=4\int \frac{\sqrt{t}}{{(t+1)}^2}dt$
Put $u=\sqrt{t}\Rightarrow du=\frac{1}{2\sqrt{t}}dt$
$\therefore I=8\int \frac{u^2}{{(u^2+1)}^2}du=8\int (\frac{2}{u^2+1}-\frac{1}{{(u^2+1)}^2})du$
$=\frac{4\left((u^2+1){\tan }^{-1}u-u\right)}{u^2+1}+C$
$=\frac{4\left((t+1){\tan }^{-1}\left(\sqrt{t}\right)-\sqrt{t}\right)}{t+1}+C$
$=\sqrt{\frac{2+x}{2-x}}(x-2)+4{\tan }^{-1}\left(\sqrt{\frac{2+x}{2-x}}\right)+C$
Therefore
${\int }_0^2\sqrt{\frac{2+x}{2-x}}dx={[\sqrt{\frac{2+x}{2-x}}(x-2)+4{\tan }^{-1}\left(\sqrt{\frac{2+x}{2-x}}\right)]}_0^2$
$=[-2+4\frac{\pi }{2}-]-0=2\pi -2$