5.3 The Fundamental Theorem of Calculus/7

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\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}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \text{Therefore, } g'(x)=\frac{1}{x^3+1}}