How many solutions has the following equation? Determine how many real solutions has the following equation:
$\sqrt[5]{2x+1+2\sqrt{{x^2+1}}}+\sqrt[5]{2x+1-2\sqrt{{x^2+1}}} = 2$
My first idea was to use the graph for the square root function: f:C->D, f(x)= $\sqrt[a]{b}$, if a is odd then f(x) is increasing on R so it should have only one solution but it doesn't 'feel' right. 
Thanks in advance.

Let $f(x)$ be a function defined by the expression.
Notice that the second term has a value of $0$ when $x=\frac{3}{4}$ and the first term has a value of $\sqrt[5]{5}$ which is less than $2$.
The derivative of $f(x)$ is
$$ \frac{1}{5}(2x+1+2\sqrt{x^2+1})^{-4/5}\left(2+\frac{2x}{\sqrt{x^2+1}}\right)+ \frac{1}{5}(2x+1-2\sqrt{x^2+1})^{-4/5}\left(2-\frac{2x}{\sqrt{x^2+1}}\right)$$
This $f^\prime(x)$ is positive for all $x>\frac{3}{4}$ so the function is increasing on the interval $\left(\frac{3}{4},\infty\right)$ and has a value smaller than $2$ at $x=\frac{3}{4}$. 
Notice that $f(n^5)>n$, so the expression has no upper bound.
Thus, since the function is continuous and is increasing for $x>\frac{3}{4}$ and has a value less than $2$ at $x=\frac{3}{4}$, and has no upper bound, there is only one solution to the equation.
