Evaluate:

By doing long division of the given equation we get

Quotient $=x+3$

Remainder $=-4$

$\therefore$ We can write the above equation as

$\Rightarrow x+3-\frac{4}{x+2}$

$\therefore$ The above equation becomes

$\Rightarrow \int \mathrm{x}+3-\frac{4}{\mathrm{x}+2} \mathrm{dx}$

$\Rightarrow \int \mathrm{x} \mathrm{dx}+3 \int \mathrm{dx}-4 \int \frac{1}{\mathrm{x}+2} \mathrm{dx}$

We know $\int x d x=\frac{x^{n}}{n+1} ; \int \frac{1}{x} d x=\ln x$

$\Rightarrow \frac{x^{2}}{2}+3 x-4 \ln (x+2)+c .$ (Where $c$ is some arbitrary constant)