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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.

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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. $(1)$.

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  • $\begingroup$ Don't think so @RainiervanEs. $\endgroup$ – Dmoreno Aug 24 '14 at 11:16
  • $\begingroup$ quite frankly, no. I'll get $\frac{13x}{3(3-2x)}+\frac{2}{3}$. How did you get the square free polynomial? $\endgroup$ – monoid Aug 24 '14 at 11:46
  • $\begingroup$ Hi again @monoid. It seems you have been able to solve your problem. Sorry for my late response. Glad to help! Indeed, when you have fractions where the degree of the numerator is the same or greater than the degree of the denominator, it's always convenient to perform polynomial division, as you have noticed. Cheers! $\endgroup$ – Dmoreno Aug 24 '14 at 19:19
  • $\begingroup$ I tried polynomial division but did not get far with my attempt. If you could spare some time would you show me the steps how you proceeded? $\endgroup$ – monoid Aug 24 '14 at 19:28
  • $\begingroup$ Sure, there you have it. $\endgroup$ – Dmoreno Aug 24 '14 at 19:37
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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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