JEE Main 201912 Apr, Morning shiftMathematicsEasyIntegral Calculus369× since 2002The integral ∫2x3−1x4+xdx\int {{{2{x^3} - 1} \over {{x^4} + x}}} dx∫x4+x2x3−1dx is equal to : (Here C is a constant of integration)AOption A. loge∣x3+1∣x2+C{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + Clogex2∣x3+1∣+CBOption B. 12loge∣x3+1∣x2+C{1 \over 2}{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C21logex2∣x3+1∣+CCOption C. loge∣x3+1x∣+C{\log _e}\left| {{{{x^3} + 1} \over x}} \right| + Clogexx3+1+CDOption D. 12loge(x3+1)2∣x3∣+C{1 \over 2}{\log _e}{{{{\left( {{x^3} + 1} \right)}^2}} \over {\left| {{x^3}} \right|}} + C21loge∣x3∣(x3+1)2+CCheckSkip · show solution
