Evaluate the integral $\int \sqrt{x^2-2x-3}dx$

Question

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?

Answer 1

Write $x^2 - 2x - 3 = (x-1)^2 - 4 $, thus

$$ \int \sqrt{ (x-1)^2 - 2^2 } dx = \int \sqrt{u^2 - 2^2 } du $$

Let $u = 2 \sec \theta $, then ....

Answer 2

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 $\; I=\dfrac12t\sqrt{t^2-4}-2\operatorname{argch}\Bigl(\dfrac t2\Bigr)$.