$I=\int \frac{1}{\left(a^2-b^2 x^2\right)^{3 / 2}} d x$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$I =\int \frac{1}{\left(a^2-b^2 x^2\right)^{3 / 2}}=\frac{-1}{2 a^2} \int \frac{-2 a^2}{x^3\left(\frac{a^2}{x^2}-b^2\right)^{3 / 2}} d x$
$\Rightarrow I =\frac{-1}{2 a^2} \int\left(\frac{a^2}{x^2}-b^2\right)^{-3 / 2} d\left(\frac{a^2}{x^2}-b^2\right)$
$\Rightarrow I =\frac{1}{a^2}\left(\frac{a^2}{x^2}-b^2\right)^{-1 / 2}+C=\frac{x}{a^2\left(a^2-b^2 x^2\right)^{1 / 2}}+C$