Q.Evaluate: $\displaystyle\int \sqrt{x^2 - 4x + 6}\,dx$.

Completing the square, $x^2-4x+6=(x-2)^2+2$, then applying the standard formula $\int\sqrt{u^2+a^2}\,du$ gives the result.

Complete the square inside the root: $x^2-4x+6 = (x-2)^2+2$.

Let $u=x-2$, so $du=dx$, and $a^2=2$ ($a=\sqrt2$).

$\displaystyle\int\sqrt{x^2-4x+6}\,dx = \int\sqrt{u^2+a^2}\,du$

Using the standard formula $\displaystyle\int\sqrt{u^2+a^2}\,du = \frac{u}{2}\sqrt{u^2+a^2}+\frac{a^2}{2}\ln\left|u+\sqrt{u^2+a^2}\right|+c$:

$= \dfrac{u}{2}\sqrt{u^2+2} + \dfrac{2}{2}\ln\left|u+\sqrt{u^2+2}\right|+c$

Substituting back $u=x-2$:

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