y = ∫ 1 − 3 x 1 u 3 1 + u 2 d u {\displaystyle y=\int _{1-3x}^{1}{\frac {u^{3}}{1+u^{2}}}du}
d d x ( y ) = d d x ( ∫ 1 − 3 x 1 u 3 1 + u 2 d u ) = ( 0 ) ⋅ ( 1 ) 3 1 + ( 1 ) 2 − ( − 3 ) ⋅ ( 1 − 3 x ) 3 1 + ( 1 − 3 x ) 2 {\displaystyle {\frac {d}{dx}}(y)={\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}}}}
Therefore, y ′ = 3 ( 1 − 3 x ) 3 1 + ( 1 − 3 x ) 2 {\displaystyle {\text{Therefore, }}y'={\frac {3(1-3x)^{3}}{1+(1-3x)^{2}}}}
