Question:
$\sqrt{x^{2}+4 x+1}$
Solution:
Let $I=\int \sqrt{x^{2}+4 x+1} d x$
$=\int \sqrt{\left(x^{2}+4 x+4\right)-3} d x$
$=\int \sqrt{(x+2)^{2}-(\sqrt{3})^{2}} d x$
It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+\mathrm{C}$
$\therefore I=\frac{(x+2)}{2} \sqrt{x^{2}+4 x+1}-\frac{3}{2} \log \left|(x+2)+\sqrt{x^{2}+4 x+1}\right|+C$