Evaluate:

By doing long division of the given equation we get

Quotient $=\frac{x}{3}+\frac{1}{9}$

Remainder $=\frac{43}{9}$

$\therefore$ We can write the above equation as

$\Rightarrow \frac{x}{3}+\frac{1}{9}+\frac{43}{9}\left(\frac{1}{3 x+2}\right)$

$\Rightarrow \frac{x}{3}+\frac{1}{9}+\frac{43}{9}\left(\frac{1}{3 x+2}\right)$

$\therefore$ The above equation becomes

$\Rightarrow \int \frac{x}{3}+\frac{1}{9}+\frac{43}{9}\left(\frac{1}{3 x+2}\right) d x$

$\Rightarrow \frac{1}{3} \int x d x+\frac{1}{9} \int d x+\frac{43}{9} \int \frac{1}{3 x+2} d x$

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

$\Rightarrow \frac{1}{3} \times \frac{x^{2}}{2}+\frac{1}{9} \times \frac{x^{2}}{2}+\frac{43}{9} \ln (3 x+2)+c$

$\Rightarrow \frac{x^{3}}{6}+\frac{x^{2}}{18}+\frac{43}{9} \ln (3 x+2)+c \cdot($ Where $c$ is some arbitrary constant)