Integration by Substitution of Fraction involving e 
Find $\int\frac{2}{e^{2x}+4}$ using $u=e^{2x}+4$

The answer is $\frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c$
I must have made a mistake somewhere as my answer is not the same. Apologies the question may be too specific, but I am teaching myself calculus.
$\int\frac{2}{e^{2x}+4}$ 
let $u = e^{2x} +4$
$\frac{dy}{dx}=2e^{2x}$
$dx=\frac{1}{2}e^{-2x}du$
$u = e^{2x} +4$
$e^{2x} = u-4$
$e^{-2x} = \frac{1}{u-4}$
$dx=\frac{1}{2}(\frac{1}{u-4})du$
Hence the integral is:
$$
\begin{aligned}
&\int\frac{2}{e^{2x}+4} \\
=& 2\int\frac{1}{u}\frac{1}{2}\left(\frac{1}{u-4}\right)du\\
=&\int\frac{1}{u}\left(\frac{1}{u-4}\right)du\\
=&\int\frac{1}{u^2-4u}du\\
=&\int u^{-2}-\frac{1}{4}u^{-1}du\\
=&\frac{u^{-1}}{-1}-\frac{1}{4}\ln(u)+c\\
=&-\frac{1}{e^{2x}+4}-\frac{1}{4}\ln(e^{2x}+4)+c\\
=&-e^{-2x}-\frac{1}{4}-\frac{1}{4}\ln(e^{2x}+4)+c\\
\ne& \frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c\\
\end{aligned}
$$
???? help

You have made a mistake: 
Note that  $$\int\frac{1}{u^2-4u}du \ne \int u^{-2}-\frac{1}{4}u^{-1}du$$
By the way 
$$\int\frac{2dx}{e^{2x}+4} =     \int\frac{2e^{2x}dx}{e^{2x}(e^{2x}+4)}  = $$
$$\int\frac{du}{u(u+4)}$$
Where $$u= e^{2x}$$
is a short way to go.
