What is the easiest way to integrate a simple isothermal equation of state?

[Peregrine]

In my continued pursuit of isothermal equation of state solutions, I've come upon a very simple integral I can't recall how to do, and was looking for assistance.

It is:

$$du = \int \frac{a}{T^{0.5}v(v+b)}dv$$

where a, T and b are constants. so:

$$du = \frac{a}{T^{0.5}} \int \frac{1}{v(v+b)}dv$$

I tried doing a u-substitution for v(v+b), but that leaves dv = du/(2v+b) which isn't permissible. Also, I tried integration by parts, using u=1/v and v = 1/(v+b), but that gets really messy with an integration of ln at the end. Does anyone have an easier way to do this simple integral that I'm missing? Thank you for the help.

 

[Pseudo Statistic]

The trick with these integrals is to split it up into "partial fractions", or whatever.
Like, in your case, you want to find an A and B such that:
$$= \dfrac{1}{ v(v+b) } = \dfrac{A}{v} + \dfrac{B}{v+b}$$
Think you can go from there?

 

[Peregrine]

I think I have it - this means that:
$$\frac {1}{v} \frac {1}{v+b} = \frac {A(v+b)+Bv}{v(v+b)}$$

This would imply, since the denominators are the same, that:
$$1 = A(v+b)+Bv.$$

I can then solve for A and B by choosing easy values for v. So, if v=0, then A=1/b. If v=-b, then B=-1/b. Plugging that back into the original equation leads to:

$$du = \frac{a}{T^{0.5}} \int \frac{1}{bv}-\frac{1}{b(v+b)}dv$$

This is much easier to integrate, so:

$$u2 - u1 = \frac{a}{bT}[(ln(v) - ln (v+b)]_2 - \frac{a}{bT}[ln (v) - ln (v+b)]_1$$

simplifying to:$$u2 - u1 = [\frac{a}{bT} ln \frac {v}{(v+b)}]_2 - [\frac{a}{bT}ln \frac {v}{(v+b)}]_1$$

I think that's it? Thanks for the help!