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 ?   
 A: 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..
A: 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: $$\int \frac{du}{3+2 u} = \frac{1}{2} \ln|3+2u|+C= \frac{1}{2} \ln|3+2 \tan x|+C.$$
A: $$\int\dfrac{\color{red}{\sec^2(x)}}{\color{blue}{3+ 2 \tan(x)}}\,dx$$
This is actually an easy u-substitution problem. Look at the denominator and the numerator. Note that:
$$\frac{d}{dx}(\color{blue}{3+ 2 \tan(x)})=2\color{red}{\sec^2 (x)}$$
Cool, and then the $2$ is easy to get rid of because it's just a constant. Although this is easy to solve in your mind, let's do u-substituion. We'll set the denominator to be $u$:
$$\begin{align} u&=\color{blue}{3+2\tan(x)} \\du&=2\color{red}{\sec^2(x)}\; dx \\&\dots \\ \color{red}{\sec^2x} &=\frac{du}{2dx} \\ \color{blue}{3+2\tan(x)} &= u \end{align}$$
Substitute:
$$\require{cancel} \int\dfrac{\color{red}{\sec^2(x)}}{\color{blue}{3+ 2 \tan(x)}}\,dx= \int \dfrac{\frac{du}{2\cancel{dx}}}{u}\;\cancel{dx}=\int \dfrac{du}{2u}=\frac12 \int \frac{du}{u}$$
What is this? Well, remember this:
$$\frac{d}{dx} \ln u = \frac1{u} \frac{du}{dx}$$
So the integral of $\frac{du}{u}$ is very simply $\ln u$.
Our answer is therefore:
$$\frac12 \int \frac{du}{u}=\frac12\ln \left|\underbrace{u}_{\color{blue}{3+2\tan(x)}} \right|=\begin{equation}\boxed{\;\therefore \frac12 \ln \left|3+2\tan(x)\right|+C \;}\end{equation}$$
