Question

If $\int \frac{dx}{(1-x)\sqrt{1-x^2}}=K\sqrt{\frac{x+1}{1-x}}+C$ then K is equal to

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $1$

Let $I=\int \frac{dx}{(1-x)\sqrt{1-x^2}}=\int \frac{dx}{{(1-x)}^2\sqrt{\frac{1+x}{1-x}}}$
Substitute $\frac{1+x}{1-x}=t\Rightarrow \frac{2}{{(1-x)}^2}dx=dt$
$\therefore I=\frac{1}{2}\int \frac{1}{\sqrt{t}}dt=\sqrt{t}+c=\sqrt{\frac{x+1}{1-x}}+C$
$\Rightarrow K=1$