Evaluate the following integrals:

Let $I=\int(x+1) e^{x} \log \left(x e^{x}\right) d x$

$\mathrm{Xe}^{\mathrm{x}}=\mathrm{t}$

$\left(1 \times e^{x}+x e^{x}\right) d x=d t$

$(x+1) e^{x} d x=d t$

$I=\int \log t d t$

$=\int 1 \times \log t d t$

Using integration by parts,

$=\log \mathrm{t} \int \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}} \log \mathrm{t} \int \mathrm{dt}$

$=\mathrm{t} \log \mathrm{t}-\int \frac{1}{\mathrm{t}} \mathrm{t} \mathrm{dt}$

$=\mathrm{tlog} \mathrm{t}-\mathrm{t}+\mathrm{c}$

$=\mathrm{t}(\log \mathrm{t}-1)+\mathrm{c}$

Substitute value for $t$,

$I=x e^{x}\left(\log x e^{x}-1\right)+c$