Find the values of $x$ such that $2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ is independent of $x$. 
Find the values of $x$ such that 
  $$2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$$ is independent of $x$.

Checking for $x\in [-1,1]$
In the taken domain $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ comes out to be $2\tan^{-1}x$ hence the taken function comes out to be equal to $4\tan^{-1}x$ hence the function is clearly dependent on $x$.
Now checking for $x\in (1,\infty)$
In the taken domain $2\tan^{-1}x$ comes out to be $\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ and hence the net sum becomes independent of $x$. 
Now checking for $x\in (-\infty,-1)$
In the taken domain $2\tan^{-1}x$ comes out to be $-\pi-\sin^{-1}\left(\frac{2x}{1+x^2}\right)$  and hence the net sum becomes $-\pi$ therefore becomes, independent of $x$.
But the answer has been mentioned as just $x\in [1,\infty)$ Can anybody tell me why the second set has not been included.
 A: Differentiate it:
$$\left(2\arctan x+\arcsin\frac{2x}{1+x^2}\right)'=\frac2{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac1{\sqrt{1-\frac{4x^2}{(1+x^2)^2}}}=$$
$$=\frac2{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac{1+x^2}{\sqrt{(1-x^2)^2}}=\frac2{1+x^2}+\frac{2(1-x^2)}{(1+x^2)|1-x^2|}$$
We can thus see that if $\;|1-x^2|=-(1-x^2)\iff 1-x^2<0\iff |x|>1\;$ , then the above last expression is zero and thus the function is a constant one and doesn't depend on $\;x\;$ .
A: The author of the question may have been thinking purely geometrically, picturing the situation below, with $x$ the (non-negative) length of a segment:

For completeness, here's the argument one can make ...
Writing $\alpha := \angle BAC = \angle BAD$ and $\beta := \angle CBD$, we have
$$\tan \alpha = x \qquad\text{and}\qquad \sin\beta = \frac{2x}{1+x^2} = \sin 2\alpha$$
(the latter by an aspect of the Law of Sines). 
Certainly, $\alpha = \operatorname{atan}{x}$ for all $x$. On the other hand, $\operatorname{asin}\left(\frac{2x}{1+x^2}\right)$ is either $\beta$ or $2\alpha$, depending upon which is non-obtuse; this condition depends upon how $\alpha$ compares to $\pi/4$, which in turn depends upon how $x$ compares to $1$. Therefore, noting that opposite angles in a cyclic quadrilateral are supplementary, we can write:
$$2 \operatorname{atan} x + \operatorname{asin}\left(\frac{2x}{1+x^2}\right) \;=\; \begin{cases}
2\alpha + \phantom{2}\beta = \pi &\text{, for } x \geq 1 \\[4pt]
2\alpha + 2\alpha = 4 \operatorname{atan}x &\text{, otherwise}
\end{cases}$$

Without an explicit restriction on the sign of $x$, though, you're correct: the solution to the problem stated should be $|x| \geq 1$, including negative values of $x$.
Edit. Strictly speaking, the answer cannot be $|x| \geq 1$, because then the sign of the sum in question would depend upon the sign of $x$. Perhaps the most accurate way to describe the result is:


*

*The sum is constant (with value $\pi$) for $x \geq 1$.

*The sum is constant (with value $-\pi$) for $x \leq -1$.

*The sum is non-constant (with value $4\operatorname{atan}x$) for $-1< x < 1$.

A: Let $\tan^{-1}x=y\implies-\dfrac\pi2\le y\le\dfrac\pi2\iff-\pi\le2y\le\pi$
$$\implies\dfrac{2x}{1+x^2}=\sin2y$$
$$\sin^{-1}\dfrac{2x}{1+x^2}=\begin{cases}-2y-\pi &\mbox{if }2y<-\dfrac\pi2\iff x<-1 \\ 
2y & \mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\\ \pi-2y & \mbox{if }2y>\dfrac\pi2\end{cases}$$
Can you take it home from here?
See also: Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
