$\begin{aligned}
& \text {Let } \mathrm{I}=\int\left(x^{3 x}+x^{2 m}+x^m\right)\left(2 x^{2 m}+3 x m+6\right)^{\frac{1}{m}} d x \\
& =\int\left(x^{3 m}+x^{2 m}+x^m\right)\left(\left(2 x^{2 m}+3 x^m+6\right) \frac{x^m}{x^m}\right)^{\frac{1}{m}} d x \\
& =\int\left(x^{3 m-1}+x^{2 m-1}+x^{m-1}\right)\left(2 x^{2 m}+3 x^{2 m}+6 x^m\right)^{\frac{1}{m}} d x
\end{aligned}$
Put $2 x^{2 m}+3 x^{2 m}+6 x^m=t$
$\begin{aligned}
& \Rightarrow\left(6 m x^{3 m-1}+6 m x^{2 m-1}+6 m x^{m-1}\right) d x=d t \\
& \Rightarrow\left(x^{3 m-1}+x^{2 m-1}+x^{m-1}\right) d x=\frac{d t}{6 m} \\
& \therefore \mathrm{I}=\frac{1}{6 m} \int t^{\frac{1}{m}} d t=\frac{1}{6 m} \cdot \frac{t^{\frac{1}{m}}+1}{\left(\frac{1}{m}+1\right)}+c \\
& \Rightarrow \mathrm{I}=\frac{\left(2 x^{3 m}+3 x^{2 m}+6 x^m\right)}{6(m+1)}+c
\end{aligned}$