For \(x \gt 3\), \(\int\sqrt{x^2-2x-3}\,dx\) is equal to (C is a constant of integration)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{x-1}{2}\sqrt{x^2-2x-3}-2\log\left|x-1+\sqrt{x^2-2x-3}\right|+C\)
Completing the square:
\( x^2-2x-3 = (x-1)^2-2^2 \)
With u = x − 1 and a = 2, we use the standard integral
\( \displaystyle \int\sqrt{u^2-a^2}\,du = \frac{u}{2}\sqrt{u^2-a^2}-\frac{a^2}{2}\log\left|u+\sqrt{u^2-a^2}\right|+C \)
Here \(\frac{a^2}{2} = 2\), so
\( \displaystyle \int\sqrt{x^2-2x-3}\,dx = \frac{x-1}{2}\sqrt{x^2-2x-3}-2\log\left|x-1+\sqrt{x^2-2x-3}\right|+C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{x-1}{2}\sqrt{x^2-2x-3}+2\log\left|x-1+\sqrt{x^2-2x-3}\right|+C\) — Incorrect. The plus sign belongs to \(\int\sqrt{u^2+a^2}\,du\); for \(u^2-a^2\) the log term is subtracted.
Option (3) \(\displaystyle\frac{x-1}{2}\sqrt{x^2-2x-3}-4\log\left|x-1+\sqrt{x^2-2x-3}\right|+C\) — Incorrect. The log coefficient is \(\frac{a^2}{2} = 2\), not \(a^2 = 4\).
Option (4) \(\displaystyle\frac{x}{2}\sqrt{x^2-2x-3}-2\log\left|x-1+\sqrt{x^2-2x-3}\right|+C\) — Incorrect. The first term needs u = x − 1, not x, in front of the root.