∫ 1 x 1 t 3 + 1 d t {\displaystyle \int _{1}^{x}{\frac {1}{t^{3}+1}}dt} g ′ ( x ) = ( 1 ) ( 1 1 + x 3 ) {\displaystyle g'(x)=(1)\left({\frac {1}{1+x^{3}}}\right)}
