5.3 The Fundamental Theorem of Calculus/17

FTC #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 G(x)=f^\prime(x)} or in other words ${\frac {d}{dx}}\left[\int \limits _{a(x)}^{b(x)}F(x)dx\right]$ is $\ b^{\prime }(x)*f(b(x))-a^{\prime }(x)*f(a(x))$

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

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

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

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

or simplified to 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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