Question

If $\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}=\frac{A\sqrt{x}}{\sqrt{1-x}}+\frac{B}{\sqrt{1-x}}+C$, where C is a real constant then A + B =

• A. 3
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (B)

0

Take $u=\sqrt{x}\Rightarrow du=\frac{dx}{2\sqrt{x}}$

So,

$=\int \frac{2udu}{(1+u)u\sqrt{1-u^2}}$

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

Put $u=\cos 2\theta$, $du=-2\sin 2\theta d\theta$

$=\int \frac{-2^2\sin 2\theta d\theta }{(1+\cos 2\theta )\sin 2\theta }$

$=\frac{-4}{2}\int {\sec }^2\theta d\theta$

$=-2\tan \theta +C$

$=-2\sqrt{\frac{1-cos2\theta }{1+cos2\theta }}+C$

$=-2\sqrt{\frac{1-u}{1+u}}+C$

$=-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$

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

$\therefore A=-2\&B=2$

$\therefore A+B=0$