∫√(x^2 - 4x + 5) dx =

The correct answer is Option (2) → \(\left(\displaystyle\frac{x - 2}{2}\right)\sqrt{x^2 - 4x + 5} + \displaystyle\frac{1}{2}\log\left|(x - 2) + \sqrt{x^2 - 4x + 5}\right| + C\)

Completing the square under the root:

    \(x^2 - 4x + 5 = (x - 2)^2 + 1\)

By 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 t = x − 2 and a = 1, so \(\frac{a^2}{2} = \frac{1}{2}\):

    \( \displaystyle \int\sqrt{x^2 - 4x + 5}\,dx = \left(\frac{x - 2}{2}\right)\sqrt{x^2 - 4x + 5} + \frac{1}{2}\log\left|(x - 2) + \sqrt{x^2 - 4x + 5}\right| + C \)

All four options share the first term; only the coefficient \(\frac{a^2}{2} = \frac{1}{2}\) of the logarithm decides the answer.