Evaluate the following integrals:

Let $I=\int(x+1) \log x d x$

Using integration by parts,

$=\log x \int(x+1) d x-\int \frac{d}{d x} \log x \int(x+1) d x$

We know that, $\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}=\frac{1}{\mathrm{x}}$

$=\log x\left(\frac{x^{2}}{2}+x\right)-\int \frac{1}{x}\left(\frac{x^{2}}{2}+x\right) d x$

$=\left(\frac{x^{2}}{2}+x\right) \log x-\int \frac{x}{2} d x-\int d x$

$=\left(\frac{x^{2}}{2}+x\right) \log x-\frac{x^{2}}{4}-x+c$

$=\left(\frac{x^{2}}{2}+x\right) \log x-\left(\frac{x^{2}}{4}+x\right)+c$