let g(x)be the inverse of the function f(x)and f''(x)= 1/1+x3. then g''(x)is equal to

F”(x) = 1/1+x^3
So integrating we get f’(x) = 1/6 log (x^2-x+1) +1/3 log (x+1) +1/ root 3 tan inverse of{ (2x-1)/root 3}
Similarly we integrate this and find f(x)
Then find inverse of f(x) as g(x)
Then differentiate g(x) two times to get g”(x)
Regards,
Nirmal Singh
Askiitians Faculty