Evaluate the following integrals:

Assume $(x+\log x)=t$

$\mathrm{d}(x+\log x)=\mathrm{dt}$

$\Rightarrow\left(1+\frac{1}{\mathrm{x}}\right) \mathrm{dx}=\mathrm{dt}$

$\Rightarrow \frac{x+1}{x} d x=d t$

Substituting $\mathrm{t}$ and $\mathrm{dt}$

$\Rightarrow \int t^{2} d t$

$\Rightarrow \frac{t^{3}}{3}+c$

But $t=x+\log x$

$\Rightarrow \frac{(x+\log x)^{3}}{3}+c$