Using $\sec(x)$ for integral.

The difference between the two answers is:

\begin{align} &\quad\frac {2x\color{red}{+}8}{\sqrt{x^2+4x}}-\frac {8}{x+\sqrt{x^2+4x}} \\&=\frac {(2x+8)(x+\sqrt{x^2+4x}) - 8\sqrt{x^2+4x}}{(x+\sqrt{x^2+4x})\sqrt{x^2+4x}} \\&=\frac {2x^2+8x+2x\sqrt{x^2+4x}} {(x+\sqrt{x^2+4x})\sqrt{x^2+4x}} \\&=\frac {2(x^2+4x+x\sqrt{x^2+4x})} {x\sqrt{x^2+4x}+x^2+4x} \\&=2 \end{align}

which is a constant. So both answers are correct.

Hint:

Another way:

For $x>0,$

$$I=\int\dfrac{\sqrt{x^2+4x}}{x^2}\ dx=\int\dfrac{\sqrt{x+4}}{x^{3/2}}\ dx$$

Integrating by parts,

$$I=\sqrt{x+4}\int\dfrac{dx}{x^{3/2}}-\int\left(\dfrac{d\sqrt{x+4}}{dx}\int\dfrac{dx}{x^{3/2}}\right)dx$$

$$=-\sqrt{x+4}\cdot\dfrac{2}{\sqrt x}+\int\dfrac{dx}{\sqrt{(x+2)^2-2^2}}$$

Now observe that $\dfrac8{\sqrt{x^2+4x}+x}=\dfrac{8(\sqrt{x^2+4x}-x)}{4x}=2\sqrt{\dfrac{x+4}x}-2$