Question

The integral $\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$ is equal to (where C is a constant of integration):

• A. $-2\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}+C$
• B. $-\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$
• C. $-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$
• D. $2\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}+C$

Answer 1

The correct option is C $-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$

$I=\int \frac{dx}{(1+\sqrt{x})\cdot \sqrt{x}\sqrt{1-x}}$
Put $1+\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}dx=dt$
$\Rightarrow I=\int \frac{2dt}{t\sqrt{2t-t^2}}$
Again put $t=\frac{1}{z}\Rightarrow dt=\frac{-1}{z^2}dz\Rightarrow I=2\int \frac{-\frac{1}{z^2}dz}{\frac{1}{z}\sqrt{\frac{2}{z}-\frac{1}{z^2}}}=2\int \frac{-dz}{\sqrt{2z-1}}=-2\sqrt{2z-1}+c$
$=-2\sqrt{\frac{2}{t}-1}+c=-2\sqrt{\frac{2-t}{t}}+c=-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+c$