Question

$\int \frac{dx}{x^2\sqrt{4-x^2}}$ is equal to

• A. $\frac{1}{4}\left(\frac{\sqrt{4-x^2}}{x}\right)+C$
• B. $\frac{1}{2}\left(\frac{\sqrt{4-x^2}}{x}\right)+C$
• C. $-\frac{1}{4}\left(\frac{\sqrt{4-x^2}}{x}\right)+C$
• D. $-\frac{1}{2}\left(\frac{\sqrt{4-x^2}}{x}\right)+C$

Answer 1

Answer: (C)

$-\frac{1}{4}\left(\frac{\sqrt{4-x^2}}{x}\right)+C$

$\int \frac{dx}{x^2\sqrt{4-x^2}}isequaltoletx=2\sin \theta dx=2\cos \theta .d\theta =\int \frac{2\cos \theta .d\theta }{4{\sin }^2\theta .2\cos \theta }$
Upon Canceling, $=\frac{1}{4}\int {cosec}^2\theta .d\theta =\frac{1}{4}(-\cot \theta )+C=-\frac{1}{4x}\sqrt{4-x^2}+C$