If \(\int \mathrm{f}(x) \mathrm{d} x=\psi(x)\), then \(\int x^5 \mathrm{f}\left(x^3\right) \mathrm{d} x\) is equal to

\(\int \mathrm{f}(x) \mathrm{d} x=\psi(x)\)
Consider
\(\mathrm{I} =\int x^5 \mathrm{f}\left(x^3\right) \mathrm{d} x \)
\( =\int x^3 \cdot x^2 \mathrm{f}\left(x^3\right) d x \)
Let \(x^3=\mathrm{t}\)
\(\Rightarrow 3 x^2 \mathrm{~d} x=\mathrm{dt}\)
\(\therefore =\int \frac{t}{3} \mathrm{f}(\mathrm{t}) \mathrm{dt} \)
\( =\frac{1}{3} \int \mathrm{t} f(\mathrm{t}) \mathrm{dt} \)
\( =\frac{1}{3}\left[\mathrm{t} \int \mathrm{f}(\mathrm{t}) \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t}) \cdot \int \mathrm{f}(\mathrm{t}) \mathrm{dt}\right] \)
\( =\frac{1}{3}\left[\mathrm{t} \cdot \psi(\mathrm{t})-\int \psi(\mathrm{t}) \mathrm{dt}\right]+\mathrm{c} \)
\( =\frac{1}{3}\left(x^3 \psi\left(x^3\right)\right)-\int \psi\left(x^3\right) x^2 \cdot \mathrm{~d} x+\mathrm{c}\)