5.3 The Fundamental Theorem of Calculus/17

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

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

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

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

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

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