Solve $ 1 + \frac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \frac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $ 
Solve for $x \in \mathbb{R}$
$$ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $$

I tried some substitutions and squaring but that didn't help. I also tried to use inequalities as done in my previous problem, but that too didn't help.
 A: Given : $ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}}  {1+\sqrt{2-2x}} $      
Let $\alpha = x+3 $ ; $\beta = 2x+2$ ; $\left(\beta - \alpha\right) = x-1$  
$\implies 1+ \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = x + \dfrac{\sqrt{\beta}}  {1+\sqrt{4-\beta}} $  
$\implies \alpha + \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = \beta + \dfrac{\sqrt{\beta}}  {1+\sqrt{4-\beta}} $  
Let $f(x) = x + \dfrac{\sqrt{x}}{1+\sqrt{4-x}}$, the given equation becomes,  
$f(\alpha) = f(\beta)$  
Note that $f(x)$ is monotonic increasing in its domain,  
$\therefore \alpha = \beta$  
$\implies x+3 = 2x +2$  
$\implies x=\boxed{1}$
A: I hope that my solution is simpler:
\begin{eqnarray}
&&1+\frac{\sqrt{x+3}}{1+\sqrt{1-x}}=x+\frac{\sqrt{2x+2}}{1+\sqrt{2-2x}}\\
&\Longleftrightarrow& x-1+\frac{\sqrt{2x+2}}{1+\sqrt{2-2x}}-\frac{\sqrt{x+3}}
{1+\sqrt{1-x}}=0\\
&\Longleftrightarrow& x-1+\frac{\sqrt{2x+2}+\sqrt{2x+2}\sqrt{1-x}-\sqrt{x+3}-\sqrt{x+3}\sqrt{2-2x}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}=0\\
&\Longleftrightarrow& x-1+\frac{\frac{x-1}{\sqrt{2x+2}+\sqrt{x+3}}+\frac{4(x-1)}{\sqrt{2x+2}\sqrt{1-x}+\sqrt{x+3}\sqrt{2-2x}}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}=0\\
&\Longleftrightarrow& (x-1)\left(1+\frac{\frac{1}{\sqrt{2x+2}+\sqrt{x+3}}+\frac{4}{\sqrt{2x+2}\sqrt{1-x}+\sqrt{x+3}\sqrt{2-2x}}}{(1+\sqrt{2-2x})(1+\sqrt{1-x})}\right)=0\\
&\Longleftrightarrow& x=1.
\end{eqnarray}
