Using $\sec(x)$ for integral. 
Find the undefined integral $\int \frac{\sqrt{x^2+4x}}{x^2}\mathrm{dx}$

$$\displaystyle\int \dfrac{\displaystyle\sqrt{x^2+4x}}{x^2}\mathrm{dx}$$
I tried to rearrange the square root and I got:
$$\sqrt{(x+2)^2-4}$$ and I substitute the $x+2$ with $2\sec(u)$ so indeed I got these two:
$$x+2=2\sec(u) \\
\sqrt{x^2+4x}=2\tan(u) \\
\mathrm{dx}=2\tan(u)\sec(u)\mathrm{du}$$
And the integral turns out like this:
$$\displaystyle \int \dfrac{\tan^2(u)\sec(u)}{(\sec(u)-1)^2}\mathrm{du}$$
And I continued to rearrange the integral:
$$\displaystyle\int \dfrac{1-\cos^2(u)}{\cos(u)(1-\cos(u))^2}\mathrm{du}$$
I apply partial fractions method, saying $\cos(u)=u$ without integral sign and I ended up with:
$$\displaystyle\int \dfrac{1}{\cos(u)}\mathrm{du}+2\displaystyle\int \dfrac{1}{1-\cos(u)}\mathrm{du}$$
One can easily integrate those integrals se I skip the calculating part. In the end I get:
$$\ln\left|\sec(u)+\tan(u)\right|+2(-\cot(u)-\csc(u))+\mathrm{C}$$
I drew a triangle:
Knowing $\sec(u)=\dfrac{x+2}{2}$ says us $\cos(u)=\dfrac{2}{x+2}$

As I calculate I got:
$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{2x-8}{\sqrt{x^2+4x}}+\mathrm{C}$$
Bu the answer key is:
$$\ln\left|x+2+\sqrt{x^2+4x}\right|-\dfrac{8}{x+\sqrt{x^2+4x}}+\mathrm{C}$$
I have been thinking where my wrong is for five hours but I couldn't find anything. Please if you see any gap tell me. Thanks.

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.
