MathematicsMedium64×since 2004Q4014If ∫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 toA3x2{3 \over {{x^2}}}x23B−16x3- {1 \over {6{x^3}}}−6x31C−12x3- {1 \over {2{x^3}}}−2x31D−12x2- {1 \over {2{x^2}}}−2x21Check answerSkip