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.

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