JEE Main 20198 Apr, Evening shiftMathematicsMediumIntegral Calculus369× since 2002If ∫dxx3(1+x6)2/3=xf(x)(1+x6)13+C\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C}∫x3(1+x6)2/3dx=xf(x)(1+x6)31+C where C is a constant of integration, then the function ƒ(x) is equal toAOption A. 3x2{3 \over {{x^2}}}x23BOption B. −16x3- {1 \over {6{x^3}}}−6x31COption C. −12x3- {1 \over {2{x^3}}}−2x31DOption D. −12x2- {1 \over {2{x^2}}}−2x21CheckSkip · show solution
