Evaluate the following integrals:
$\int(x+1) e^{x} \log \left(x e^{x}\right) d x$
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 $t$,
$I=x e^{x}\left(\log x e^{x}-1\right)+c$
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