Q.Evaluate ∫ dx / (x² - 4x + 13).

Integration by Completing the Square

You can integrate $\dfrac{1}{x^2+1}$ or $\dfrac{1}{x^2-a^2}$ on sight. But a quadratic denominator such as $x^2+4x+5$ fits neither standard form directly. The fix is to rewrite the quadratic as a perfect square plus (or minus) a constant, turning it into a form you already know.

The move

For $x^2+bx+c$, add and subtract $\left(\dfrac{b}{2}\right)^2$:

After substituting $u=x+\dfrac{b}{2}$ the integral collapses to $\displaystyle\int\frac{du}{u^2+k^2}$ or $\displaystyle\int\frac{du}{u^2-k^2}$.

Case A — leads to inverse tangent

Complete the square: $x^2+4x+5=(x+2)^2+1$. With $u=x+2$,

Case B — leads to a logarithm

Here $x^2-6x+5=(x-3)^2-4$. With $u=x-3$ the denominator $u^2-4$ factors, so use partial fractions:

$\int\frac{du}{u^2-4}=\frac14\log\left|\frac{u-2}{u+2}\right|+C=\frac14\log\left|\frac{x-5}{x-1}\right|+C.$ …