so I have this integral:
$$\int\frac{1}{1-v^{2}}dv$$
Now I know this isn't in the form of any of the common integrals for the trig functions, so I decided to use trig sub:
Let $$v = sin(\theta), dv = cos(\theta)$$
After substituting, I end up with $$\int sec(\theta)d\theta$$
which yields $$\ln \bigg|\sec(\theta)+tan(\theta)\bigg |$$
Right? Now if I draw a triangle, since $$\sin(\theta) = \frac{v}{1}$$ then the following substitution is made: $$\ln\bigg|\frac{v+1}{\sqrt{1-v^{2}}}\bigg |.$$ So I'm not sure how to go from this to this:
$$\frac{1}{2}\ln\bigg |\frac{v+1}{v-1}\bigg |$$
I tried solving this integral using Partial Fractions and it was very easy to get to this form, but I'm not sure where to go about with the answer I arrived at. Some help would be appreciated.
*Also, please ignore the fact that I did not add a + C, this is just the LHS of a differential equation I am trying to solve.
Thanks
Just note that $$\frac{v+1}{\sqrt{1-v^2}} = \frac{v+1}{\sqrt{1+v}\sqrt{1-v}} = \frac{\sqrt{1+v}}{\sqrt{1-v}}$$
So the two are infact equivalent under appropriate domain.