How to find the following integral? $\int\tfrac{\sec^2(x)}{3+ 2 \tan(x)}\,\mathrm dx $ Please show me how to find the following integral: $$\int\dfrac{\sec^2(x)}{3+ 2 \tan(x)}\,\mathrm dx $$
I think the solution will be something about the natural logarithm right ?   

Recognize that $\frac{d}{dx} \tan x = \sec^2 x$, and then substitute $u = \tan x$, $du = \sec^2 x\, dx$.
Using that $\int \frac{dx}{x} = \ln |x|+C$, you next obtain: $$\int \frac{du}{3+2 u} = \frac{1}{2} \ln|3+2u|+C= \frac{1}{2} \ln|3+2 \tan x|+C.$$
