<s>[INST] 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 $\dfrac{x(x^2-5)}{x^2-4}$
 [/INST] 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:
$$\ln\sqrt{\frac{e^{x^2}}{x^2-4}}+C=\ln\sqrt{\frac{e^{x^2}}{e^x+2}}$$
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