Question

Integrate the function $\sqrt{x^2+4x+6}$

Answer 1

$x^2+4x+6=(x+2{)}^2+2=(x+2{)}^2+(\sqrt{2}{)}^2$
Let $x+2=t\Rightarrow dx=dt$
Also we know that, $\int \sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\log |x+\sqrt{x^2+a^2}|+C$
$\therefore \int \sqrt{x^2+4x+6}dx=\int \sqrt{(x+2{)}^2+(\sqrt{2}{)}^2}dx$
$=\int \sqrt{t^2+(\sqrt{2}{)}^2}dt=\frac{t}{2}\sqrt{t^2+2}+\frac{2}{2}\log (t+\sqrt{t^2+2})+C$
$=\frac{x+2}{2}\sqrt{x^2+4x+6}+\log |x+2+\sqrt{x^2+4x+6}|+C$