Question

$I=\int x\sqrt{\frac{1-x}{1+x}}dx$

• A. $(\frac{x}{2}-1)\sqrt{1-x^2}-\frac{1}{2}{\sin }^{-1}x+C$
• B. $(\frac{x}{2}+1)\sqrt{1-x^2}-\frac{1}{2}{\sin }^{-1}x+C$
• C. $(\frac{x}{2}+1)\sqrt{1-x^2}-\frac{1}{2}{\cos }^{-1}x+C$
• D. $(\frac{x}{2}-1)\sqrt{1-x^2}-\frac{1}{2}{\cos }^{-1}x+C$

Answer 1

The correct option is D $(\frac{x}{2}-1)\sqrt{1-x^2}-\frac{1}{2}{\cos }^{-1}x+C$

Let $I=\int x\sqrt{\frac{1-x}{1+x}}dx$

Put $u=\frac{1-x}{1+x}\Rightarrow du=(-\frac{1-x}{{(x+1)}^2}-\frac{1}{x+1})dx$

$\therefore I=-2\int \frac{(u-1)\sqrt{u}}{{(-u-1)}^3}du$

Put $t=\sqrt{u}\Rightarrow dt=\frac{1}{2\sqrt{u}}du$

$\therefore I=4\int \frac{t^2(t^2-1)}{{(t^2+1)}^3}dt$

$=4\int \frac{1}{t^2+1}dt-12\int \frac{1}{{(t^2-1)}^2}dt+8\int \frac{1}{{(t^2-18)}^3}dt$

$=\frac{(t^2+1)(\frac{25}{{(t^2+1)}^2}+{\sin }^{-1}t-3\sin \left(2{\tan }^{-1}t\right))+3t}{t^2+1}$

$=(\frac{x}{2}-1)\sqrt{1-x^2}-\frac{1}{2}{\cos }^{-1}x+C$