$\begin{aligned} & \text { Let } \mathrm{I}=\int x^5 \mathrm{e}^{-4 x^3} \mathrm{~d} x \\ & \begin{aligned} \text { Put } x^3=\mathrm{t} \Rightarrow 3 x^2 \mathrm{~d} x & =\mathrm{dt} \\ \therefore \quad \mathrm{I}=\frac{1}{3} \int \mathrm{te}^{-4 \mathrm{t}} \mathrm{dt} & =\frac{1}{3}\left(\mathrm{t} \cdot \frac{\mathrm{e}^{-4 \mathrm{t}}}{-4}-\int 1 \cdot \frac{\mathrm{e}^{-4 \mathrm{t}}}{-4} \mathrm{dt}\right) \\ & =\frac{1}{3}\left(\frac{-\mathrm{te}^{-4 \mathrm{t}}}{4}+\frac{1}{4} \cdot \frac{\mathrm{e}^{-4 \mathrm{t}}}{-4}\right)+\mathrm{c} \\ & =\frac{1}{48} \mathrm{e}^{-4 \mathrm{t}}(-4 \mathrm{t}-1)+\mathrm{c} \\ & =\frac{1}{48} \mathrm{e}^{-4 x^3}\left(-4 x^3-1\right)+\mathrm{c}\end{aligned}\end{aligned}$
$\therefore \quad \mathrm{f}(x)=-4 x^3-1$