If ${\tan }^{-1}\frac{1}{1+2x}+{\tan }^{-1}\frac{1}{4x+1}={\tan }^{-1}\frac{2}{x^2}$  then $\text{'}x\text{'}=$

${\tan }^{-1}\left[\frac{\frac{1}{1+2x}+\frac{1}{4x+1}}{1-\frac{1}{1+2x}.\frac{1}{4x+1}}\right]={\tan }^{-1}\left(\frac{2}{x^2}\right)$

$\Rightarrow {\tan }^{-1}\left[\frac{4x+1+2x+1}{\left(4x+1\right)\left(1+2x\right)-1}\right]={\tan }^{-1}\left(\frac{2}{x^2}\right)$

$\Rightarrow {\tan }^{-1}\left[\frac{6x+2}{4x+2x+8x^2+1-1}\right]={\tan }^{-1}\left[\frac{2}{x^2}\right]$

$\Rightarrow {\tan }^{-1}\left[\frac{6x+2}{6x+8x^2}\right]={\tan }^{-1}\left[\frac{2}{x^2}\right]\Rightarrow \frac{3x+1}{3x+4x^2}=\frac{2}{x^2}$

$\Rightarrow 3x^3+x^2=6x+8x^2\Rightarrow 3x^3-7x^2-6x=0$

$\Rightarrow x\left(3x^2-7x-6\right)=0\Rightarrow x\left(3x^2-9x+2x-6\right)=0$

$x=0\left(or\right)3x\left(x-3\right)+2\left(x-3\right)=0\Rightarrow x=0\left(or\right)\left(x-3\right)\left(3x+2\right)=0$

$x=0\left(or\right)3\left(or\right)-\frac{2}{3}$