<s>[INST] 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.
 [/INST] 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 $\;x\;$ .
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