$\smallint \,\,\frac{{{x^2} - \,\,2}}{{{x^3}\,\,\sqrt {{x^2} - \,\,1} }}\,\,dx$ is equal to

$\frac{{{x^2}}}{{\sqrt {{x^2} - \,\,1} }}\,\, + \,\,C$

$ - \,\,\frac{{{x^2}}}{{\sqrt {{x^2} - \,\,1} }} + \,\,C$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

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

is equal to

View answer and explanation

Correct option: D

- \,\,\frac{{\sqrt {{x^2} - \,\,1} }}{{{x^2}}}\,\, + C

Option D is correct because after simplifying and using trigonometric substitution, the integral evaluates to

- \frac{{\sqrt {{x^2} - \,1} }}{{{{x^2}}}}\, + C