Evaluate the integral $\int \sqrt{x^2-2x-3}dx$ I'm asked to evaluate this integral: $\int \sqrt{x^2-2x-3} dx$
I don't see any other way to solve this except by trigonometric substitution, which is precisely what I did once I completed the square and got $\sqrt{4-(x-1)^2}$ as the integrand.
I then performed a substitution with $(x-1) = 2\sin(\theta)$.
But all of the answer choices contain some natural logarithms in them. I have no clue how that's possible.
What other methods are there for solving this problem?

Another method
Integrate by parts: setting $t=x-1$, we come down to the integral $\;I=\displaystyle\int\sqrt{t^2-4}\,\mathrm d t$.
Set $u=\sqrt{t^2-4}$, $\mathrm dv=\mathrm dt$, whence 
$\mathrm du=\dfrac{t\,\mathrm dt}{\sqrt{t^2-4}}, \enspace v=t$, and
\begin{align}
I&=t\sqrt{t^2-4}-\int\dfrac{t^2\,\mathrm dt}{\sqrt{t^2-4}}=t\sqrt{t^2-4}-\int\dfrac{(t^2-4)\,\mathrm dt}{\sqrt{t^2-4}}-4\int\dfrac{\mathrm dt}{\sqrt{t^2-4}} \\
&=t\sqrt{t^2-4}-I-4\operatorname{argch}\Bigl(\frac t2\Bigr)
\end{align}
so that $\; I=\dfrac12t\sqrt{t^2-4}-2\operatorname{argch}\Bigl(\dfrac t2\Bigr)$.
