Integrate the function: $\sqrt{x^{2}+4x-5}$

Let $I = \int \sqrt{x^{2}+4x-5} \, dx$
First,complete the square for the quadratic expression inside the square root:
$x^{2}+4x-5 = (x^{2}+4x+4) - 4 - 5 = (x+2)^{2} - 9 = (x+2)^{2} - (3)^{2}$
Thus,$I = \int \sqrt{(x+2)^{2} - (3)^{2}} \, dx$
Using the standard integration 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 $a=3$:
$I = \frac{(x+2)}{2} \sqrt{(x+2)^{2} - (3)^{2}} - \frac{3^{2}}{2} \ln |(x+2) + \sqrt{(x+2)^{2} - (3)^{2}}| + C$
Simplifying the expression:
$I = \frac{(x+2)}{2} \sqrt{x^{2}+4x-5} - \frac{9}{2} \ln |(x+2) + \sqrt{x^{2}+4x-5}| + C$
Where $C$ is an arbitrary constant.