Evaluate the following integrals:

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

That is,

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

Using integration by parts,

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

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

$\frac{x}{x+1}=1-\frac{1}{x+1}$

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

$=x \log (x+1)-x+\log (x+1)+c$