Inverse Function (and WolframAlpha gives different Result) I wanted to calculate the inverse function of 
$$
 f(x) = \frac{1}{x} + \frac{1}{x-1}
$$
Quite simple I thought, put
$$
 y = \frac{1}{x} + \frac{1}{x-1} = \frac{2x-1}{x(x-1)}
$$
rearrange and solve
$$
 y(x(x-1)) - 2x + 1 = 0
$$
which give the quadratic equation
$$
 yx^2 - (y + 2)x + 1 = 0
$$
Using the Solution Formula we got
$$
 x = \frac{(y+2) \pm \sqrt{y^2+4}}{2y}
$$
So the inverse function is
$$
 f^{-1}(x) = \frac{(x+2) \pm \sqrt{x^2+4}}{2x}
$$
Just to confirm I put in WolframAlpha and it gives me
$$
 \frac{-x-2}{2x} \pm \frac{\sqrt{x^2+4}}{2x}
$$
(just click on the link to start WolframAlpha with this parameter), which
is different up to a sign in the first summand, can not see an error, do you (or is WolframAlpha wrong...)?
EDIT: If the link is not working for you:


Your error is in the solution formula. You have $(y+2)^2 - 4\cdot y\cdot 1 = y^2+4 \neq (y+2)(y-2)$. It would be $y^2-4 = (y+2)(y-2)$.
