Q.Evaluate : $\int \dfrac{1}{\sqrt{x^2 + 6x + 13}}\, dx$.

Write $x^2+6x+13 = (x+3)^2 + 4$; then it is the standard integral $\int \dfrac{dt}{\sqrt{t^2+a^2}} = \log|t + \sqrt{t^2+a^2}|$.

Step 1 — complete the square.

$x^2 + 6x + 13 = (x^2 + 6x + 9) + 4 = (x+3)^2 + 2^2.$

Step 2 — substitute $t = x + 3$ (so $dt = dx$):

$\int \frac{dx}{\sqrt{x^2+6x+13}} = \int \frac{dt}{\sqrt{t^2 + 2^2}}.$

Step 3 — apply the standard result $\displaystyle\int \frac{dt}{\sqrt{t^2 + a^2}} = \log\left|t + \sqrt{t^2 + a^2}\right| + c$: …