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 1 + x 3 ) = 1 1 + x 3 {\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}{1+x^{3}}}\right)={\frac {1}{1+x^{3}}}}
![{\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}{1+x^{3}}}\right)={\frac {1}{1+x^{3}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5cf560a961e2c0da347690da04642fcc62a95391)