JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 6)
The integral $$\int {{{2{x^3} - 1} \over {{x^4} + x}}} dx$$ is equal to :
(Here C is a constant of integration)
(Here C is a constant of integration)
$${\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C$$
$${1 \over 2}{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C$$
$${\log _e}\left| {{{{x^3} + 1} \over x}} \right| + C$$
$${1 \over 2}{\log _e}{{{{\left( {{x^3} + 1} \right)}^2}} \over {\left| {{x^3}} \right|}} + C$$
Explanation
$$\int {{{2{x^3} - 1} \over {{x^4} + x}}dx = \int {{{2x - {x^{ - 2}}} \over {{x^2} + {x^{ - 1}}}}dx = \ln ({x^2} + {x^{ - 1}}} } ) + c$$
$$ \Rightarrow \ln ({x^3} + 1) - \ln x + c$$
$$ \Rightarrow \ln ({x^3} + 1) - \ln x + c$$
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