$\smallint \frac{{{x^4} - \,\,1}}{{{x^2}{{\left( {{x^4} + {x^2} + 1} \right)}^{1/2}}}}\;dx$ is equal to

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

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

$x{\left( {{x^4} + {x^2} + 1} \right)^{3/2}} + c$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Hard

1 of 1

Analyzing +4 | -1

\int \frac{{{x^4} - \,\,1}}{{{x^2}{{\left( {{x^4} + {x^2} + 1} \right)}^{1/2}}}}\;dx

is equal to

View answer and explanation

Correct option: D

\frac{{\sqrt {{x^4} + {x^2} + 1} }}{x} + c

The correct answer is D (

\frac{{\sqrt {{{{x^4} + {x^2} + 1}}} }}{x} + c

) because the integral simplifies to the derivative of the given expression.