$$\frac{\sqrt{1-x} + \frac{1}{\sqrt{1+x}}}{1 + \frac{1}{\sqrt{1-x}}}$$
I've had a go at putting everything over a common denominator in the form of:
$$\frac{\frac{\sqrt{1-x}\sqrt{1+x}+1}{\sqrt{1+x}}}{\frac{\sqrt{1-x}+1}{\sqrt{1-x}}} = \frac{\sqrt{1-x}(\sqrt{1-x^2}+1)}{(\sqrt{1-x}+1)\sqrt{1+x}}$$
Or multiplying the complex fraction by the conjugate of its base, but I'm getting nowhere.
Extensive searching hasn't made me realize the answer either— it has me baffled.
From where you left off, to remove all radicals from the denominator, we could multiply the top and bottom by $\sqrt{1+x}(1-\sqrt{1-x})$ -- the former factor to remove the factor of $\sqrt{1+x}$, and the latter being the conjugate of $1+\sqrt{1-x}$. This would result with
\begin{align*} \frac{\sqrt{1-x}(\sqrt{1-x^2}+1)\sqrt{1+x}(1-\sqrt{1-x})}{(1+x)(1-(1-x))} &= \frac{\sqrt{1-x^2}(\sqrt{1-x^2}+1)(1-\sqrt{1-x})}{x(1+x)} \\ &= \frac{(1-x^2+\sqrt{1-x^2})(1-\sqrt{1-x})}{x(1+x)} \end{align*}
You could also expand the numerator, but I don't really think it seems "simpler".
\begin{align*} \frac{(1-x^2+\sqrt{1-x^2})(1-\sqrt{1-x})}{x(1+x)} &= \frac{1-x^2+\sqrt{1-x^2}-\sqrt{1-x}+x^2\sqrt{1-x}-(1-x)\sqrt{1+x}}{x(1+x)} \\ &= \frac{(x^2-1)\sqrt{1-x}+(x-1)\sqrt{1+x}+\sqrt{1-x}\sqrt{1+x}+1-x^2}{x(1+x)} \end{align*}
There are many different ways to write this expression.
