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

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

Substitute $$x = \frac{1}{t} \implies dx = -\frac{1}{t^2} dt$$:

$$f(t) = \int \frac{-\frac{1}{t^2} dt}{(3 + \frac{4}{t^2})\sqrt{4 - \frac{3}{t^2}}} = -\int \frac{t \, dt}{(3t^2 + 4)\sqrt{4t^2 - 3}}$$

Let $$4t^2 - 3 = z^2 \implies 8t \, dt = 2z \, dz \implies t \, dt = \frac{z \, dz}{4}$$:

$$f(z) = -\int \frac{\frac{z}{4} dz}{(3(\frac{z^2 + 3}{4}) + 4)z} = -\int \frac{dz}{3z^2 + 9 + 16} = -\int \frac{dz}{3z^2 + 25}$$

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

Substituting $$z = \sqrt{4t^2 - 3} = \sqrt{\frac{4}{x^2} - 3} = \frac{\sqrt{4 - 3x^2}}{x}$$:

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

Given $$f(0) = 0$$: $$0 = -\frac{1}{5\sqrt{3}} \left(\frac{\pi}{2}\right) + C \implies C = \frac{\pi}{10\sqrt{3}}$$

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

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

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

$$\alpha^2 + \beta^2 = 5^2 + (\sqrt{3})^2 = 25 + 3 = 28$$