$$ \begin{aligned} & f(x)=\int \frac{d x}{\left(3+4 x^2\right) \sqrt{4-3 x^2}} \\\\ & x=\frac{1}{t} \\\\ & =\int \frac{\frac{-1}{t^2} d t}{\frac{\left(3 t^2+4\right)}{t^2} \frac{\sqrt{4 t^2-3}}{t}} \\\\ & =\int \frac{-d t \cdot t}{\left(3 t^2+4\right) \sqrt{4 t^2-3}}: \text { Put } 4 t^2-3=z^2 \\\\ & =-\frac{1}{4} \int \frac{z d z}{\left(3\left(\frac{z^2+3}{4}\right)+4\right) z} \end{aligned} $$

$$ \begin{aligned} & =\int \frac{-d z}{3 z^2+25}=-\frac{1}{3} \int \frac{\mathrm{dz}}{\mathrm{z}^2+\left(\frac{5}{\sqrt{3}}\right)^2} \\\\ & =-\frac{1}{3} \frac{\sqrt{3}}{5} \tan ^{-1}\left(\frac{\sqrt{3} z}{5}\right)+C \\\\ & =-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{4 t^2-3}\right)+C \\\\ & f(x)=-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{\frac{4-3 \mathrm{x}^2}{x^2}}\right)+C \\\\ & \because \mathrm{f}(0)=0 \because \mathrm{c}=\frac{\pi}{10 \sqrt{3}} \end{aligned} $$

$$ \begin{aligned} & f(1)=-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5}\right)+\frac{\pi}{10 \sqrt{3}} \\\\ & f(1)=\frac{1}{5 \sqrt{3}} \cot ^{-1}\left(\frac{\sqrt{3}}{5}\right)=\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{5}{\sqrt{3}}\right) \\\\ & \alpha=5: \beta=\sqrt{3} \therefore \alpha^2+\beta^2=28 \end{aligned} $$