[Math] 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?

Best Answer

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 ....

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