Integrate $\displaystyle\int{\frac{1}{2+3x}dx}$ Integrate $$\int{\frac{1}{2+3x}dx}$$

What should I do for this integral?  How do I evaluate it?

Thanks!
HINT:

Put $2+3x=u$ so that $3dx=du$

and we know $$\int \frac{du}u=\ln |u|+C$$ where $C$ is an arbitrary constant for indefinite integration
Integrate 1/2+3xdx............ let u=2+3x, so that 3dx=du, .'. dx=du/3u.......and we know........int:du/u=In\u/+C, where c is the arbitrary constant for indifinite integration........ Therefore, int:du/3u=1/3 In u+c..........final answer is...........1/3In(2+3x)+C. Hope this help........ By O'john