5.3 The Fundamental Theorem of Calculus/17

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

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}(g(x))=\frac{d}{dx}\left(\int_{1-3x}^{1}\frac{u^3}{(1+u^2)} du\right) =(0)*f(1)-(-3)*f(1-3x)}

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

which is=$3*{\frac {(1-3x)^{3}}{(1+(1-3x)^{2})}}$

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

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