Is there another way to solve $\int\frac{1}{9e^v-1}dv$ that's just as efficient?

Question

I'm solving a DE on Paul's Notes, and I stopped at this integral:

$\int\frac{1}{9e^v-1}dv$

The suggested method is to multiply top and bottom by $e^{-v}$ to put it in a form with which a sub can be easily used. Fair enough.

However, if I didn't know about this trick, what is another method I could have used to solve this that's about as quick?

Answer 1

If $\displaystyle 9e^v-1=u, 9e^vdv=du\implies dv=\frac{du}{9e^v}=\frac{du}{u+1}$

$$\int\frac{dv}{9e^v-1}=\int\frac{du}{u(u+1)}$$

Now, $$\frac1{u(u+1)}=\frac{u+1-u}{u(u+1)}=\cdots$$

Answer 2

Put $s=e^v$, then $ds=e^vdv=sdv$ or $dv=\frac{ds}{s}$. The integral now is: $$\int\frac{1}{9e^v-1}dv=\int\frac{1}{(9s-1)s}ds$$ Which is easier to solve methodly.

Answer 3

Or, along the same lines, try getting rid of the exponential by writing $v = \log x$, $dv = dx/x$, so that the integral becomes $\int \frac{dx}{x(9x-1)}$ and you're faced with the same sort of partial fraction exercise. Not much different from the previous answer, but might be what I first would have tried.