1. We need to evaluate

rac{1}{(3+4x^2)\,\sqrt{4-3x^2}}

and write

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

The question says that for $|x|>0$, the antiderivative can be written in the form

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

and asks for $\alpha^2+\beta^2$.

2. Use the substitution

$$x=\frac{2}{\sqrt{3}}\sin\theta.$$

Then

$$dx=\frac{2}{\sqrt{3}}\cos\theta\,d\theta,$$

and

$$\sqrt{4-3x^2}=\sqrt{4-3\cdot \frac{4}{3}\sin^2\theta}=\sqrt{4(1-\sin^2\theta)}=2\cos\theta.$$

Also,

$$3+4x^2=3+4\cdot \frac{4}{3}\sin^2\theta=3+\frac{16}{3}\sin^2\theta =\frac{9+16\sin^2\theta}{3}.$$

So the integral becomes

$$I=\int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}} =\int \frac{\frac{2}{\sqrt{3}}\cos\theta\,d\theta}{\left(\frac{9+16\sin^2\theta}{3}\right)(2\cos\theta)}.$$

Canceling $2\cos\theta$,

$$I=\int \frac{\frac{1}{\sqrt{3}}}{\frac{9+16\sin^2\theta}{3}}\,d\theta =\sqrt{3}\int \frac{d\theta}{9+16\sin^2\theta}.$$

3. Convert this to a standard form using $t=\tan\theta$. Then

$$\sin^2\theta=\frac{t^2}{1+t^2}, \qquad d\theta=\frac{dt}{1+t^2}.$$

Hence

$$9+16\sin^2\theta=9+16\cdot \frac{t^2}{1+t^2} =\frac{9(1+t^2)+16t^2}{1+t^2} =\frac{9+25t^2}{1+t^2}.$$

Therefore,

$$I=\sqrt{3}\int \frac{1}{\frac{9+25t^2}{1+t^2}}\cdot \frac{dt}{1+t^2} =\sqrt{3}\int \frac{dt}{9+25t^2}.$$

4. Now integrate:

$$\int \frac{dt}{9+25t^2} =\frac{1}{15}\tan^{-1}\left(\frac{5t}{3}\right)+C.$$

So,

$$I=\frac{\sqrt{3}}{15}\tan^{-1}\left(\frac{5t}{3}\right)+C.$$

Since $t=\tan\theta$ and

$$x=\frac{2}{\sqrt{3}}\sin\theta,$$

we get

$$\sin\theta=\frac{\sqrt{3}x}{2},$$ $$\cos\theta=\frac{\sqrt{4-3x^2}}{2}.$$

Thus

$$\tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{\frac{\sqrt{3}x}{2}}{\frac{\sqrt{4-3x^2}}{2}} =\frac{\sqrt{3}x}{\sqrt{4-3x^2}}.$$

Hence

$$I=\frac{\sqrt{3}}{15}\tan^{-1}\left(\frac{5}{3}\cdot \frac{\sqrt{3}x}{\sqrt{4-3x^2}}\right)+C =\frac{\sqrt{3}}{15}\tan^{-1}\left(\frac{5x}{\sqrt{3}\sqrt{4-3x^2}}\right)+C.$$

5. Rewrite in the required form. Multiply numerator and denominator inside by $\sqrt{3}$:

$$\frac{5x}{\sqrt{3}\sqrt{4-3x^2}} =\frac{5\sqrt{3}x}{3\sqrt{4-3x^2}}.$$

So one valid form is

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

with

$$\alpha=\frac{\sqrt{3}}{15}, \qquad \beta=\frac{5\sqrt{3}}{3}.$$

Then

$$\alpha^2=\frac{3}{225}=\frac{1}{75}, \qquad \beta^2=\frac{25\cdot 3}{9}=\frac{25}{3}.$$

Thus

$$\alpha^2+\beta^2=\frac{1}{75}+\frac{25}{3} =\frac{1+625}{75}=\frac{626}{75},$$

which is not an integer.

So the intended form in the question is almost certainly the standard one without irrational coefficients split awkwardly. Let us instead derive by direct differentiation the cleaner representation:

Assume

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

Let

$$u=\frac{\beta x}{\sqrt{4-3x^2}}.$$

Then

$$\nu' = \beta\cdot \frac{4}{(4-3x^2)^{3/2}}.$$

Also,

$$1+\nu^2=1+\frac{\beta^2x^2}{4-3x^2} =\frac{4+(\beta^2-3)x^2}{4-3x^2}.$$

Hence

$$\frac{d}{dx}\tan^{-1}(\nu) =\frac{\nu'}{1+\nu^2} =\frac{4\beta}{\sqrt{4-3x^2}\,[4+(\beta^2-3)x^2]}.$$

Therefore,

$$f'(x)=\alpha\cdot \frac{4\beta}{\sqrt{4-3x^2}\,[4+(\beta^2-3)x^2]}.$$

We want this equal to

$$\frac{1}{(3+4x^2)\sqrt{4-3x^2}}.$$

So

$$\alpha\cdot \frac{4\beta}{4+(\beta^2-3)x^2} = \frac{1}{3+4x^2}.$$

Matching denominators gives

$$4+(\beta^2-3)x^2 = k(3+4x^2)$$

for some constant $k$, and numerator gives

$$4\alpha\beta = \frac{1}{k}.$$

From constant term,

$$4=3k \Rightarrow k=\frac{4}{3}.$$

Then from $x^2$ coefficient,

$$\beta^2-3 = 4k = \frac{16}{3} \Rightarrow \beta^2=\frac{25}{3}.$$

Also,

$$4\alpha\beta = \frac{3}{4} \Rightarrow \alpha\beta=\frac{3}{16}.$$

Using $\beta=\frac{5}{\sqrt{3}}$,

$$\alpha=\frac{3}{16}\cdot \frac{\sqrt{3}}{5}=\frac{3\sqrt{3}}{80},$$

which is inconsistent with the earlier exact antiderivative, showing that the question statement as transcribed is incomplete/misprinted regarding the exact required form of $\alpha,\beta$.

6. Since the stored answer is $28$, the intended standard answer is likely obtained from a cleaner equivalent form such as

$$f(x)=\frac{1}{4\sqrt{7}}\tan^{-1}\left(\frac{2\sqrt{7}x}{\sqrt{4-3x^2}}\right)+C,$$

or a similar misprinted variant, for which $\alpha^2+\beta^2$ can become an integer. But for the given integrand, the exact derivation does not yield $28$.

Therefore, based on the actual calculus, the stored answer appears inconsistent with the provided question text.