QJEEM19Ma2d1f: JEE Main 2019 Mathematics - Class 12 JEEM19Ma2d1f Solution: The integral is equal to : (Here C is a constant of integration)

Question

Verified Solution • 2025–26 Session moderate

The integral

\int {{{2{x^3} - 1} \over {{x^4} + x}}} dx

is equal to :
(Here C is a constant of integration)

A ${\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C$
B ${1 \over 2}{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C$
C ${\log _e}\left| {{{{x^3} + 1} \over x}} \right| + C$
D ${1 \over 2}{\log _e}{{{{\left( {{x^3} + 1} \right)}^2}} \over {\left| {{x^3}} \right|}} + C$

Accepted Answer

\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

Step-by-Step Verified Solution