Finding substitution in the integral $\int{\frac{2+3x}{3-2x}}dx$ In a problem sheet I found the integral $$\int{\frac{2+3x}{3-2x}}dx.$$
In the solution the substitution $z=3-2x$ is given which yields $x=\frac{3-z}{2}$ and $dx=-\frac{1}{2}dz$. We have
$$\int{\frac{2+3x}{3-2x}}dx = \int{\frac{2+3\left(\frac{3-z}{2}\right)}{z}}\left(-\frac{1}{2}dz\right)=\cdots$$
How do I recognise these kind of substitutions and how do I choose them?
After reading the line $z=3-2x \Rightarrow x=\frac{3-z}{2} \Rightarrow dx=-\frac{1}{2}dz$. I was able to solve the integral. My main problem is, how do I recognise from the integrand that I have to substitute and solve for the free variable $x$. Is there a trick to remember?
Dmoreno's hint which I was at first not able to follow but after some thinking made me realise that equations like $\frac{a+bx}{c-dx}$ can be alternated like:
$$\frac{a+bx}{c-dx} = \left(\frac{a+bx}{c-dx} + \frac{b}{d}\right) - \frac{b}{d} = \frac{ad+bc}{d(c-dx)} - \frac{b}{d},$$
therefore $$\int{\frac{a+bx}{c-dx}}dx = \frac{ad+bc}{d}\cdot\int{\frac{1}{c-dx}}dx - \frac{b}{d}\cdot \int dx,$$
which made my problem much easier.

Hint: note that 
$$ \frac{2+3x}{3-2x} = \frac{13}{2 (3-2x)}-\frac{3}{2} \tag{1}$$
Can you take it from here?

Addendum:
\begin{array}{cccccc}
\color{blue}{2}     & \color{blue}{+} & \color{blue}{3 x }& | & \color{red}{3} & \color{red}{-} & \color{red}{2x} \\ \hline
-9/2  & + & 3 x & | &  \color{purple}{-\frac{3}{2} } &  &     \\
\color{green}{13/2}  & + & 0   & |
\end{array}
Since $\color{blue}{D} = \color{red}{d} \, \color{purple}{c} + \color{green}{r}, $ it follows that:
$$2+ 3x = -(3-2x)\frac{3}{2} +\frac{13}{2},$$
which readily yields eq. $(1)$.
