Question:

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

Updated On: May 31, 2024

$\frac{-1}{6}(1-\frac{1}{x^2})^{\frac{3}{2}}+C$
$\frac{1}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C$
$\frac{-1}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C$
$\frac{4}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C$
$\frac{-4}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C$

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The Correct Option is B

The correct option is (B) : $\frac{1}{3}(1-\frac{1}{x^2})^{\frac{3}{2}}+C$

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