Find the solution $(1+2y)dx+(4-x^2)dy=0$ 
solve $(1+2y)dx+(4-x^2)dy=0$ using separation of variables

$$(1+2y)dx+(4-x^2)dy=0\implies (4-x^2)dy=-(1+2y)dx\implies \frac{dy}{(-1-2y)}=\frac{dx}{(4-x^2)}$$
Integrate the two sides:
$$\begin{align}
-\frac{1}{2}\ln(-2y-1)&=\int \frac{dx}{4-x^2}\\
\\&\text{now,}\\
\int \frac{dx}{4-x^2}&=\int \frac{1}{2(2-x)}+\int \frac{1}{2(2+x)}\\
&=-\frac{\ln(2-x)}{2}+\frac{\ln(2+x)}{2}\\
\\&\text{now,}\\
-\frac{1}{2}\ln(-2y-1)&=-\frac{\ln(2-x)}{2}+\frac{\ln(2+x)}{2}+C\\
\implies\ln(-2y-1)&={\ln(2-x)}-{\ln(2+x)}-2C\\
\implies e^{\ln(-2y-1)}&=e^{{\ln(2-x)}-{\ln(2+x)}-2C}\\
\implies-2y\cdot e^{-1}&=e^{{\ln(2-x)}-{\ln(2+x)}-2C}\\
\implies y&=-\frac{1}{2}\cdot e^{{\ln(2-x)}-{\ln(2+x)}-2C+1}
\end{align}$$
Is the following correct? are the steps valid?
the answer from wolfram is a bit different 

Not quite.  Note that
$$\int \frac{1}{-1-2y}\,dy=-\frac12 \log|2y+1| \tag 1$$
and
$$\begin{align}
\int \frac{1}{4-x^2}\,dx&=\int \left(\frac{1/4}{2-x}+\frac{1/4}{2+x}\right)\,dx\\\\
&=\frac14 \log\left|\frac{x+2}{x-2}\right|+C \tag 2
\end{align}$$
Then, upon equating $(1)$ and $(2)$, we arrive at
$$-\frac12 \log|2y+1|=\frac14 \log\left|\frac{x+2}{x-2}\right|+C $$
from which we obtain
$$|2y+1|=A \sqrt{\frac{|x-2|}{|x+2|}}$$
Note that for $y>-1/2$, we find that
$$y=A' \sqrt{\frac{|x-2|}{|x+2|}} - \frac12$$
