**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?