Derivative of $\ln\sqrt{\frac{e^{x^2}}{e^x+2}}$? 
Being: $\ln f(x)=\log_ef(x)$

I started derivating $$\ln\sqrt{\dfrac{e^{x^2}}{e^x+2}}$$ but I get to a point that I don't know how to follow.
I try to get it by derivating the logarithm directly and by using the logarithmic properties such us: $$\ln f(x)^n=n\ln f(x)$$ and $$\ln \frac{f(x)}{Q(x)}=\ln f(x)-\ln Q(x)$$ but I don't get it.
By the time I have done this:
\begin{align}
f(x)&=\ln\sqrt{\dfrac{e^{x^2}}{e^x+2}}\\
&=\frac{\ln e^{x^2}-\ln (e^x+2)}{2}\\
&=\frac{x^2-\ln (e^x+2)}{2}\\
\text{And derivating:}\\
f'(x)&=\frac{1}{2}\left(\frac{d}{dx}x^2-\frac{d}{dx}\ln (e^x+2)\right)\\
&=x-\frac{1}{2}\left(\frac{e^x}{e^x+2}\right)
\end{align}
So I finally get this:

$f'(x)=x-\dfrac{e^x}{2e^x+4}$

But WolframAlpha says that $$f'(x)=\frac{x(x^2-5)}{x^2-4}$$
If I'm wrong, what do I do wrong? And if I'm right, how can I get from $x-\dfrac{e^x}{2e^x+4}$ to $\dfrac{x(x^2-5)}{x^2-4}$
 A: Your calculation is perfectly correct, and efficient. The answer that you say Alpha gives is incorrect, and in particular not equivalent to yours. It is possible that Alpha interpreted what you typed as something different from what you intended. If you look carefully at what Alpha thinks it is differentiating, you might find where missing parentheses led Alpha astray.
A: A check, if you trusting your integration skills more than your derivating ones:
Integrate your $f'(x)$, and if you compute your derivative correctly, you will then get your original function, $f(x)$ with some constant. (For more reading: Fundamental theorem of calculus)
$$\int\frac{x(x^2-5)}{x^2-4}\ dx=\int (x-\frac{1}{2(x-2)}+\frac{1}{2(x+2)})\ dx = $$ 
$$=\frac{x^2}{2}-\frac{1}{2}\ln|x-2|-\frac{1}{2}\ln|x+2|+C=\frac{1}{2}(x^2-\ln|x^2-4|)+C=\frac{1}{2}{\ln{e^{x^2}}-\frac{1}{2}\ln|x^2-4|}+C=$$
$$=\frac{1}{2}\ln\frac{e^{x^2}}{x^2-4}+C=\ln\sqrt{\frac{e^{x^2}}{x^2-4}}+C$$
Now, the question, is there $C\in\mathbb{R}$, such that:
$$\ln\sqrt{\frac{e^{x^2}}{x^2-4}}+C=\ln\sqrt{\frac{e^{x^2}}{e^x+2}}$$
