g ( x ) = ∫ 1 x 1 t 3 + 1 d t {\displaystyle g(x)=\int _{1}^{x}{\frac {1}{t^{3}+1}}dt}
d d x [ g ( x ) ] = d d x [ ∫ 1 x 1 t 3 + 1 d t ] = ( 1 ) ( 1 ( x ) 3 + 1 ) − ( 0 ) ( 1 ( 1 ) 3 + 1 ) = 1 x 3 + 1 {\displaystyle {\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{1}^{x}{\frac {1}{t^{3}+1}}dt\right]=(1)\left({\frac {1}{(x)^{3}+1}}\right)-(0)\left({\frac {1}{(1)^{3}+1}}\right)={\frac {1}{x^{3}+1}}} Therefore, g ′ ( x ) = 1 x 3 + 1 {\displaystyle {\text{Therefore, }}g'(x)={\frac {1}{x^{3}+1}}}
![{\displaystyle {\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{1}^{x}{\frac {1}{t^{3}+1}}dt\right]=(1)\left({\frac {1}{(x)^{3}+1}}\right)-(0)\left({\frac {1}{(1)^{3}+1}}\right)={\frac {1}{x^{3}+1}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6cf243c7869bd94ab3b91c68f55b995a0942c27d)
