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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:

[tex] du = \int \frac{a}{T^{0.5}v(v+b)}dv[/tex]

where a, T and b are constants. so:

[tex] du = \frac{a}{T^{0.5}} \int \frac{1}{v(v+b)}dv[/tex]

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.

It is:

[tex] du = \int \frac{a}{T^{0.5}v(v+b)}dv[/tex]

where a, T and b are constants. so:

[tex] du = \frac{a}{T^{0.5}} \int \frac{1}{v(v+b)}dv[/tex]

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.

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