Define the image of the function $f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$ Question
Define the range of the function: 
$$f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$$
Answer attempt
I assume that the domain for $f(x)$ is in ${\rm I\!R}$.
If we draw a triangle with the sides: $1, x$ and $\sqrt{1+x^2}$, with the angle $\theta$ opposite of $x$, we get:
$$\sin{\frac{x}{\sqrt{1+x^2}}} \rightarrow$$ $$\theta = \arcsin{\frac{x}{\sqrt{1+x^2}}}$$
$$\tan{\frac{x}{1}} \rightarrow$$ $$\theta = \arctan{x}$$
This means that: 
$$\arctan x = \arcsin{\frac{x}{\sqrt{1+x^2}}}$$
The original expression can thus be written:
$$2 \arctan x + \arcsin \frac {2x}{1+x^2} = 2 \arcsin \frac{x}{\sqrt{1+x^2}} + \arcsin \frac {2x}{1+x^2}$$
We know that $\arcsin$ has the domain $-1 \leq x \leq 1$ and the range $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$
If we start with: $\arcsin \frac {2x}{1+x^2}$
$x=1 \rightarrow \arcsin{\frac{2}{2}} = \arcsin 1 = \frac{\pi}{2}$
$x=-1 \rightarrow \arcsin{\frac{-2}{2}} = \arcsin -1 = -\frac{\pi}{2}$
$-1 \leq x \leq 1 \rightarrow -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
And analyzing: $2\arcsin \frac{x}{\sqrt{1+x^2}}$
$x=1 \rightarrow 2\arcsin{\frac{1}{\sqrt{2}}} \rightarrow \sin{\theta} = \frac{1}{\sqrt{2}} \rightarrow \theta = \frac{\pi}{4}$
$x=-1 \rightarrow 2\arcsin{-\frac{1}{\sqrt{2}}} \rightarrow \sin{\theta} = -\frac{1}{\sqrt{2}} \rightarrow \theta = -\frac{\pi}{4}$
$-1 \leq x \leq 1 \rightarrow -2(\frac{\pi}{4}) \leq y \leq 2(\frac{\pi}{4}) \rightarrow -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
The range of $f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$ is therefore $-\pi \leq y \leq \pi$
I wonder if you feel that my reasoning is correct? Is there some better way to solve this problem?

This range is correct. While you start with the assumption that the domain for $f(x)$ is $x\in\mathbb{R}$, we know that the domain of $f(x)$ is limited by the domain of $\arcsin(x)$ which is $[-1,1]$.
So, this means that $-1\leq\dfrac{2x}{1+x^2}\leq 1$ which is precisely $x\in[-1,1]$
Since $\arctan(x)$ is an increasing function, we know that the maximum value of $f(x)$ is at $x=1$ and and the minimum at $x=-1$ If we substitute these values, we get the same range that you found.
For a more rigorous verification, we could take the derivative, but I think this is a more intuitive explanation.
