Question

Integrate the function $\sqrt{x^2+4x+1}$

Answer 1

Let $I=\int \sqrt{x^2+4x+6}dx$
$=\int \sqrt{x^2+4x+4+2}dx$
$=\int \sqrt{(x^2+4x+4)+2}dx$
$=\int \sqrt{(x+2{)}^2+(\sqrt{2}{)}^2}dx$
We know that, $\int \sqrt{x^2+a^2}dx=d\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log |x+\sqrt{x^2+a^2}|+C$
$\therefore I=\frac{(x+2)}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\log |(x+2)+\sqrt{x^2+4x+6}|+C$
$=\frac{(x+2)}{2}\sqrt{x^2+4x+6}+\log |(x+2)+\sqrt{x^2+4x+6}|+C$