5.3 The Fundamental Theorem of Calculus/17

$g(x)=\int _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du$

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}(g(x))={\frac {d}{dx}}\left(\int _{1-3x}^{1}{\frac {u^{3}}{(1+u^{2})}}du\right)=(0)\cdot {\frac {(1)^{3}}{(1+(1)^{2})}}(0)*f(1)-(-3)*f(1-3x)}

which is equal to $(3)*f(1-3x)$

which is=Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 3*{\frac {(1-3x)^{3}}{(1+(1-3x)^{2})}}}

or simplified to 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{3*(1-3x)^3}{(1+(1-3x)^2)}}

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