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 ?   

Set $$u =  3+2\tan(x)$$ then $$du =  2\sec^2(x)dx$$ this would then imply 
$$\int\dfrac{\sec^2(x)}{3+ 2 \tan(x)}\,\mathrm dx = \int\frac{\sec^2(x)dx}{u} = \int\frac{\frac{1}{2}du}{u} = \frac12\int\frac{du}{u} = \frac12\ln(u) +c $$  $$ = \frac12 \ln(3+2\tan(x)) +c $$ etc..
