If $\tan^{-1}(x-1)+\tan^{-1}(x)+\tan^{-1}(x+1)=\tan^{-1}(3x),$ then x = 

-1

$\tan^{-1}(x-1)+\tan^{-1}(x)+\tan^{-1}(x+1)=\tan^{-1}(3x)$ then $\tan^{-1}(x-1)+\tan^{-1}(x+1)=\tan^{-1}(3x)-\tan^{-1}(x)$

$⇒\tan^{-1}[\frac{(x-1)+(x+1)}{1-(x-1)+(x+1)}]=\tan^{-1}[\frac{3x-x}{1+(3x)(x)}]$

$⇒\tan^{-1}[\frac{2x}{2-x^2}]=\tan^{-1}[\frac{2x}{1+3x^2}]⇒\frac{2x}{2-x^2}=\frac{2x}{1+3x^2}$

$⇒x=0,\frac{1}{2},\frac{-1}{2}$