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∫x2−4x+5 dx=\int \sqrt{x^2 - 4x + 5} \, dx =

Solution

✅ Correct Option: 2

x2−4x+5=(x−2)2+1x^2 - 4x + 5 = (x-2)^2 + 1. Using ∫t2+a2 dt=t2t2+a2+a22log⁡∣t+t2+a2∣+C\int\sqrt{t^2 + a^2}\,dt = \frac{t}{2}\sqrt{t^2+a^2} + \frac{a^2}{2}\log|t + \sqrt{t^2+a^2}| + C with t=x−2,a=1t = x-2, a = 1: =x−22x2−4x+5+12log⁡∣(x−2)+x2−4x+5∣+C= \frac{x-2}{2}\sqrt{x^2-4x+5} + \frac{1}{2}\log|(x-2)+\sqrt{x^2-4x+5}| + C.

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