5.3 The Fundamental Theorem of Calculus/17

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

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

${\text{Therefore, }}g'(x)={\frac {3(1-3x)^{3}}{1+(1-3x)^{2}}}$