TITLE: Find the inverse function $x + \sqrt {x}$
QUESTION [1 upvotes]: $ Y = x + \sqrt {x} $ 
Hello , I want to find the inverse function of this function , I know that it's injective How to prove the $f(x) = \sqrt{x + \sqrt{x}}$ is injective. but do not know how to make the inverse function step by step.

REPLY [2 votes]: Thanks to the @egreg 's answer I found an easier solution :
add $ +\frac{1}{4} - \frac{1}{4}$ we have :
$$y = x + \sqrt{x}  +\frac{1}{4} - \frac{1}{4}$$
then :
$$y=(\sqrt{x}+\frac{1}{2})^2-\frac{1}{4}$$
$\sqrt{x} + \frac{1}{2}$ is clearly positive so :
$$(\sqrt{y+{\frac{1}{4}}}-\frac{1}{2})^2=x$$
then : 
$$x = y + \frac{1}{2}-\sqrt{y+{\frac{1}{4}}}$$
you can swap x and y to have the inverse function.

 Because the range of the function is $y ≥ 0$ we have $x ≥ 0$ for the inverse function domain.

Which is equal to this answer https://math.stackexchange.com/a/1116697/208741
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