integrate $\int \frac{1}{\sqrt{x^2 - 1}} \, dx$ I am solving an ODE

$$y^2 \, dx + \left(x\sqrt{y^2 - x^2} - xy\right)dy=0$$

I let $y=xu$ and did some algebra, and I ended up with this

$$\ln(x)+C=\int \frac{1}{\sqrt{u^2 - 1}} \, du - \int \frac{1}{u} \, du$$

I don't know how to solve the first right hand side integral. I tried to let $u^2 - 1 = t$, so it becomes

$$\int\frac{1}{\sqrt{u^2 - 1}}=\int\frac{1}{\sqrt{t}}\frac{1}{2u} \, dt = \int \frac{1}{\sqrt{t}} \frac{1}{2} \frac{1}{\sqrt{t-1}} \, dt,$$

which doesn't seem to work. The correct answer seems to be 

$$\ln \left(\sqrt{x^2 - 1} - x\right)$$

But how do we get this?
$$\int  \frac { dx }{ \sqrt { x^{ 2 }-1 }  } =\int { \frac { \sinh { t }  }{ \sinh { t }  } dt=t } +C\\ \\ x=\cosh { t } \\ dx=\sinh { t } \\ x=\frac { { e }^{ t }+{ e }^{ -t } }{ 2 } \\ 2x{ e }^{ t }={ e }^{ 2t }+1\\ { e }^{ 2t }-2x{ e }^{ t }+1=0\\ { e }^{ t }=x+\sqrt { { x }^{ 2 }-1 } $$

$$t=\ln { \left| x+\sqrt { { x }^{ 2 }-1 }  \right|  } $$
Use a substitution of $u=\sin(t)$, differentiate to get $du$ then sub into that 2nd integral.

Then evaluate that integral &amp; should progress towards the answer.