Question

$\int \frac{dx}{x^2\sqrt{4+x^2}}=$

• A. $\frac{1}{4}\sqrt{4+x^2}+c$
• B. $\frac{-1}{4}\sqrt{4+x^2}+c$
• C. $\frac{-1}{4x}\sqrt{4+x^2}+c$
• D. $\frac{9}{4x}\sqrt{4+x^2}+c$

Answer 1

The correct option is C $\frac{-1}{4x}\sqrt{4+x^2}+c$

$I=\int \frac{dx}{x^2\sqrt{4+x^2}}$

$=\int \frac{dx}{x^3\sqrt{\frac{4}{x^2}+1}}$

let $\frac{4}{x^2}+1=t$

$\frac{-8}{x^3}dx=dt$

$I=\frac{1}{-8}\int \frac{-8dx}{x^3\cdot \sqrt{\frac{4}{x^2}+1}}$

$=-\frac{1}{8}\int \frac{dt}{\sqrt{t}}$

$=\frac{-1}{8}\left[\frac{(t{)}^{-\frac{1}{2}}}{1-\frac{1}{2}}\right]=\frac{-2}{8}\sqrt{t}$

$=\frac{-1}{4}\sqrt{\frac{4}{x^2}+1}$