$\smallint \frac{{dx}}{{x\,\sqrt {{x^6}\,\, - \,\,16} }}$ is equal to

$\frac{1}{3}\,\,{\sec ^{ - 1}}\left( {\frac{{{x^3}}}{4}} \right)\,\, + \,\,c$

${\cos ^{ - 1}}\,\,\left( {\frac{{{x^3}}}{4}} \right)\,\, + \,\,c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

\int \frac{{dx}}{{x\,\sqrt {{x^6}\,\, - \,\,16} }}

is equal to

View answer and explanation

Correct option: C

\frac{1}{{12}}\,\,{\sec ^{ - 1}}\,\,\left( {\frac{{{x^3}}}{4}} \right)\,\, + \,\,c

Option C (

\frac{1}{{12}}\,{\sec ^{ - 1}}\,\,\left( {\frac{{{x^3}}}{4}} \right)\,\,+\,c

) is correct because the integral is solved using substitution and the standard integral of

\frac{1}{x\sqrt{x^2-a^2}}