<s>[INST] 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.
 [/INST] You were on the good way $$I =\int{\frac{2+3x}{3-2x}}dx$$ Change variable using $3-2x=z$, so $x=\frac{3-z}{2}$ and $dx=-\frac{1}{2} dz$ (which is what you obtained). Replace and simplify as much as you can; you should easily arrive to $$I=\int \Big(\frac{3}{4}-\frac{13}{4 z}\Big)dz=\frac{3}{4}\int dz-\frac{13}{4}\int\frac{dz}{z}$$
I am sure that you can take from here.
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