We have given that,
$\int x^3 e^{5 x} d x=\frac{e^{5 x}}{5^4}\left[\int(x)\right]+c$
By using the method of intergration by parts, we get
$\begin{aligned}
& \int x^3 e^{5 x} d x=x^3 \int e^{5 x} d x-\int\left\{\frac{d x^3}{d x} \int e^{5 x} d x\right\} d x \\
& =\frac{x^3 e^{5 x}}{5}-\int \frac{3 x^2 e^{5 x}}{5} d x+C
\end{aligned}$
By using the method of integration by parts, we get
$\begin{aligned}
& =\frac{x^3 e^{5 x}}{5}-\frac{3}{5} x^2 \int e^{5 x} d x \\
& +\frac{3}{5} \int\left\{\frac{d x^2}{d x} \int e^{5 x} d x\right\} d x+C \\
& =\frac{x^3 e^{5 x}}{5}-\frac{3 x^3 e^{5 x}}{25}+\frac{6}{25} \int x e^{5 x} d x \\
& =\frac{x^3 e^{5 x}}{5}-\frac{3 x^3 e^{5 x}}{25}+\frac{6 x}{25} \int e^{5 x} d x \\
& -\frac{6}{25} \int\left\{\frac{d x}{d x} \int e^{5 x} d x\right\} d x \\
& =\frac{x^3 e^{5 x}}{5}-\frac{3 x^3 e^{5 x}}{25}+\frac{6 x e^{5 x}}{125}-\frac{6}{125} \int e^{5 x} d x
\end{aligned}$
$\begin{aligned}
& =\frac{x^3 e^{5 x}}{5}-\frac{3 x^3 e^{5 x}}{25}+\frac{6 x e^{5 x}}{125}-\frac{6}{125}+C \\
& =\frac{e^{5 x}}{625}\left[125 x^3-75 x^2+30 x-6\right]+C
\end{aligned}$
Comparing the above equation with
$\begin{aligned}
& \frac{e^{5 x}}{5^4}[f(x)]+C, \text { we get } \\
& f(x)=5^3 x^3-75 x^2+30 x-6
\end{aligned}$