Why $\frac{x}{\sqrt{x+1}-1}$ can be written as $\sqrt{x+1}+1$? I've evaluated the first formula on $W|A$ and it says that $\sqrt{x+1}+1$ is an alternate form to the first expression. I just don't see how it's possible. 
The first thing I imagined was to write:
$$\frac{x}{\sqrt{x+1}-1}=x^1(\sqrt{x+1}-1)^{-1}$$
But it doesn't seems to give me much insight.

Note: the two functions of $x$ are NOT equal, because they have different domains.
$\frac{x}{\sqrt{x+1}-1}$ has domain $[-1,0)\cup(0,\infty)$
$\sqrt{x+1}+1$ has domain $[-1,\infty)$
Hence the former is not defined at $x=0$, while the latter is.
However, on all the points of their mutual domain, they agree; the calculation involves multiplying by the conjugate of the denominator, and may be found in the other answers.
