\(\int\sqrt{x^2-6x+13}\,dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{x-3}{2}\sqrt{x^2-6x+13}+2\log\left|x-3+\sqrt{x^2-6x+13}\right|+C\)
Completing the square:
\( x^2-6x+13 = (x-3)^2+2^2 \)
With \(t = x-3\), this is the standard integral
\( \displaystyle \int\sqrt{t^2+a^2}\,dt = \frac{t}{2}\sqrt{t^2+a^2}+\frac{a^2}{2}\log\left|t+\sqrt{t^2+a^2}\right|+C \)
With a = 2, \(\frac{a^2}{2} = 2\):
\( \displaystyle \int\sqrt{x^2-6x+13}\,dx = \frac{x-3}{2}\sqrt{x^2-6x+13}+2\log\left|x-3+\sqrt{x^2-6x+13}\right|+C \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{x-3}{2}\sqrt{x^2-6x+13}+4\log\left|x-3+\sqrt{x^2-6x+13}\right|+C\) — Incorrect. The log term carries \(\frac{a^2}{2} = 2\), not \(a^2 = 4\).
Option (3) \(\displaystyle\frac{x-3}{2}\sqrt{x^2-6x+13}-2\log\left|x-3+\sqrt{x^2-6x+13}\right|+C\) — Incorrect. The minus sign belongs to \(\int\sqrt{t^2-a^2}\,dt\), not to \(\int\sqrt{t^2+a^2}\,dt\).
Option (4) \((x-3)\sqrt{x^2-6x+13}+2\log\left|x-3+\sqrt{x^2-6x+13}\right|+C\) — Incorrect. The first term has the factor \(\frac{t}{2}\), not t.