Let $f\left(x\right)=\int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}},\left|x\right|<\frac{2}{\sqrt{3}}.$ If $f\left(0\right)=0$ and $f\left(1\right)=\frac{1}{\alpha \beta }{\tan }^{-1}\left(\frac{\alpha }{\beta }\right).\alpha ,\beta >0,$ then ${\alpha }^2-{\beta }^2$ is equal to _____

$f\left(x\right)=\int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}}$

$x=\frac{1}{t}=\int \frac{\frac{-1}{t^2}dt}{\frac{(3t^2+4)}{t^2}\frac{\sqrt{4t^2-3}}{t}}=\int \frac{-dt.t}{(3t^2+4)\sqrt{4t^2-3}}:Put4t^2-3=z^2$

$=-\frac{1}{4}\int \frac{zdz}{(3\left(\frac{z^2+3}{4}\right)+4)z}=\int \frac{-dz}{3z^2+25}=-\frac{1}{3}\int \frac{dz}{z^2+{\left(\frac{5}{\sqrt{3}}\right)}^2}=-\frac{1}{3}\frac{\sqrt{3}}{5}{\tan }^{-1}\left(\frac{\sqrt{3}z}{5}\right)+C$

$=-\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}}{5}\sqrt{4t^2-3}\right)+C$

$f\left(x\right)=-\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}}{5}\sqrt{\frac{4-3x^2}{x^2}}\right)+C$

$\because f\left(0\right)=0\because c=\frac{\pi }{10\sqrt{3}}$

$f\left(1\right)=-\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{\sqrt{3}}{5}\right)+\frac{\pi }{10\sqrt{3}}$

$f\left(1\right)=\frac{1}{5\sqrt{3}}{\cot }^{-1}\left(\frac{\sqrt{3}}{5}\right)=\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{5}{\sqrt{3}}\right)$

$\alpha =5:\beta =\sqrt{3}=\therefore {\alpha }^2-{\beta }^2=22$