5.3 The Fundamental Theorem of Calculus/7

$g(x)=\int _{1}^{x}{\frac {1}{t^{3}+1}}dt$

${\frac {d}{dx}}\left[g(x)\right]={\frac {d}{dx}}\left[\int _{1}^{x}{\frac {1}{t^{3}+1}}dt\right]=(1)\left({\frac {1}{(x)^{3}+1}}\right)-(0)\left({\frac {1}{(1)^{3}+1}}\right)={\frac {1}{x^{3}+1}}$

${\text{Therefore, }}g'(x)={\frac {1}{x^{3}+1}}$