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

Let $I = \int \sqrt{x^{2}+4x+1} \, dx$.
First,complete the square for the quadratic expression:
$x^{2}+4x+1 = (x^{2}+4x+4)-3 = (x+2)^{2}-(\sqrt{3})^{2}$.
So,$I = \int \sqrt{(x+2)^{2}-(\sqrt{3})^{2}} \, dx$.
Using the standard integral formula $\int \sqrt{x^{2}-a^{2}} \, dx = \frac{x}{2} \sqrt{x^{2}-a^{2}} - \frac{a^{2}}{2} \ln |x + \sqrt{x^{2}-a^{2}}| + C$,where $x$ is replaced by $(x+2)$ and $a = \sqrt{3}$:
$I = \frac{(x+2)}{2} \sqrt{(x+2)^{2}-(\sqrt{3})^{2}} - \frac{3}{2} \ln |(x+2) + \sqrt{(x+2)^{2}-(\sqrt{3})^{2}}| + C$.
Simplifying the expression back to the original form:
$I = \frac{(x+2)}{2} \sqrt{x^{2}+4x+1} - \frac{3}{2} \ln |(x+2) + \sqrt{x^{2}+4x+1}| + C$,where $C$ is an arbitrary constant.