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?
 A: 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.
A: Since $-1\le 2x/(1+x^2)\le 1$ for every $x\in\mathbb{R}$, the domain of $f$ is indeed the whole real axis. Moreover
$$
\lim_{x\to-\infty}f(x)=-\pi,\qquad
\lim_{x\to\infty}f(x)=\pi
$$
Let's compute the derivative:
$$
f'(x)=\frac{2}{1+x^2}+\frac{1}{\sqrt{1-\frac{4x^2}{(1+x^2)^2}}}
\frac{2(1+x^2)-4x^2}{(1+x^2)^2}
$$
A simple computation brings this to
$$
f'(x)=\frac{2}{1+x^2}\left(1+\frac{1-x^2}{|1-x^2|}\right)
$$
(which is valid for $x\ne\pm1$).
First step.
$$
\sqrt{1-\frac{4x^2}{(1+x^2)^2}}=
\sqrt{\frac{(1+x^2)^2-4x^2}{(1+x^2)^2}}=
\sqrt{\frac{1-2x^2+x^4}{(1+x^2)^2}}=
\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}=\frac{|1-x^2|}{1+x^2}
$$
Second step
$$
\frac{2(1+x^2)-4x^2}{(1+x^2)^2}=2\frac{1-x^2}{(1+x^2)^2}
$$
Third step
$$
f'(x)=\frac{2}{1+x^2}+2\frac{1+x^2}{|1-x^2|}\frac{1-x^2}{(1+x^2)^2}=
\frac{2}{1+x^2}\left(1+\frac{1-x^2}{|1-x^2|}\right)
$$
Fourth step
If $1-x^2<0$ (that is, $|x|>1$), we have $\dfrac{1-x^2}{|1-x^2|}=-1$
Fifth step
If $1-x^2>0$ (that is, $|x|<1$), we have $\dfrac{1-x^2}{|1-x^2|}=1$
Hence we have
$$
f'(x)=\begin{cases}
0 & \text{if $x<-1$}\\[4px]
\dfrac{4}{1+x^2} & \text{if $-1<x<1$}\\[4px]
0 & \text{if $x>1$}
\end{cases}
$$
Therefore we have
$$
f(x)=\begin{cases}
-\pi & \text{if $x<-1$}\\[4px]
4\arctan x & \text{if $-1\le x\le 1$}\\[4px]
\pi & \text{if $x>1$}
\end{cases}
$$

