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Step-by-Step Solution
Step 1: Understand the problem
We are given the function
f(x) = \int \frac{dx}{\bigl(3 + 4x^2\bigr)\,\sqrt{4 - 3x^2}} for \lvert x \rvert < \frac{2}{\sqrt{3}} ,
with the condition f(0)=0 . We also know
f(1) = \frac{1}{\alpha\beta}\,\tan^{-1}\bigl(\frac{\alpha}{\beta}\bigr), \ \alpha,\beta>0.
We must determine the value of \alpha^2 + \beta^2.
Step 2: Plan a suitable substitution for the integral
The integral has two parts in the denominator:
1) 3 + 4x^2
2) \sqrt{4 - 3x^2} .
A common approach for integrals of the form
\displaystyle \int \frac{dx}{\bigl(a + b\,x^2\bigr)\,\sqrt{c - d\,x^2}}
is to use trigonometric or reciprocal substitutions that simplify the expression under the square root and the polynomial factor.
Step 3: Sketch of the integral simplification
A typical sequence might involve:
• Substituting x = \frac{1}{t} (reciprocal substitution) to transform the denominator.
• Then accounting for the square root \sqrt{4 - 3x^2} with another substitution so that it simplifies to an expression in t .
Eventually, the integral simplifies to a form
\displaystyle -\int \frac{dz}{3z^2 + 25} ,
which is recognizable as an arctangent-type integral.
Step 4: Evaluate the simpler arctangent integral
We use the standard result:
\int \frac{dz}{z^2 + a^2} \;=\; \frac{1}{a}\,\tan^{-1} \Bigl(\frac{z}{a}\Bigr).
In our transformed integral, there is a factor of 3 in the denominator, so we write:
-\int \frac{dz}{3z^2 + 25} \;=\; -\int \frac{dz}{3\Bigl(z^2 + \frac{25}{3}\Bigr)}.
Factor out the 3 :
= -\frac{1}{3}\,\int \frac{dz}{z^2 + \frac{25}{3}}.
Let a^2 = \frac{25}{3} so a = \frac{5}{\sqrt{3}} . This yields:
-\frac{1}{3} \times \frac{\sqrt{3}}{5}\,\tan^{-1}\Bigl(\frac{\sqrt{3}\,z}{5}\Bigr)
\;=\; -\frac{1}{5\sqrt{3}}\,\tan^{-1}\Bigl(\frac{\sqrt{3}\,z}{5}\Bigr).
Afterward, we back-substitute z in terms of the original x .
Step 5: Determine the integration constant using f(0) = 0
When we transform back completely to x , we get an expression for f(x) plus some constant C .
We use the condition f(0)=0 to solve for C . Evaluating the limit as x \to 0 helps pin down the exact expression for f(x) .
Step 6: Find f(1)
Substituting x=1 into the final form of f(x) gives:
f(1) = \;-\frac{1}{5\sqrt{3}}\,\tan^{-1}\Bigl(\frac{\sqrt{3}}{5}\Bigr)
\;+\;\text{(a constant term)}.
After simplifying (and carefully evaluating the constant from Step 5), one obtains a neatly reduced expression typically in the form:
f(1) = \frac{1}{5\sqrt{3}}\,\tan^{-1}\Bigl(\frac{5}{\sqrt{3}}\Bigr).
This is given to match the form
\frac{1}{\alpha \beta}\,\tan^{-1}\Bigl(\frac{\alpha}{\beta}\Bigr).
Step 7: Identify \alpha and \beta , then compute \alpha^2 + \beta^2
By comparing
\tfrac{1}{5\sqrt{3}}\tan^{-1}\bigl(\tfrac{5}{\sqrt{3}}\bigr)
with
\tfrac{1}{\alpha \beta}\tan^{-1}\bigl(\tfrac{\alpha}{\beta}\bigr),
it is clear that
\alpha = 5
and
\beta = \sqrt{3}.
Hence,
\alpha^2 + \beta^2 \;=\; 5^2 + (\sqrt{3})^2
\;=\; 25 + 3
\;=\; 28.
Final Answer
The value of \alpha^2 + \beta^2 is \boxed{28} .
