Evaluate the following integrals:

Let I $=\int \frac{\log (x+2)}{(x+2)^{2}} \mathrm{dx}$

$\frac{1}{x+2}=t$

$\frac{-1}{(x+2)^{2}} d x=d t$

$I=-\int \log \left(\frac{1}{t}\right) d t$

Using integration by parts,

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

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

We know that, $\frac{\mathrm{d}}{\mathrm{dt}} \log \mathrm{t}=\frac{1}{\mathrm{t}}$ and $\int \mathrm{dt}=\mathrm{t}$

$I=\log t \int d t-\int\left(\frac{d}{d t} \log t \int d t\right) d t$

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

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

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

Replace the value of $t$,

$=2 e^{\sqrt{x}}(\sqrt{x}-1)+c$