Key Concepts and Formulas

• Trigonometric Substitution: Using trigonometric functions to simplify integrals involving square roots of quadratic expressions.
• Inverse Trigonometric Functions: Understanding the relationship between $\tan^{-1}(x)$ and $\cot^{-1}(x)$, specifically $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$.
• Indefinite Integration Rules: $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

Step-by-Step Solution

Step 1: Rewrite the integral. We are given the integral $f(x) = \int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}}$.

Step 2: Perform the substitution $x = \frac{1}{t}$. Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2} dt$. Substituting into the integral gives:

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

Step 3: Perform the substitution $z^2 = 4t^2 - 3$. Let $z^2 = 4t^2 - 3$. Then $2z dz = 8t dt$, so $t dt = \frac{1}{4} z dz$. Also, $t^2 = \frac{z^2 + 3}{4}$. Substituting into the integral yields:

$f(x) = \int \frac{-\frac{1}{4} z dz}{\left(3\left(\frac{z^2 + 3}{4}\right) + 4\right) z} = -\frac{1}{4} \int \frac{dz}{\frac{3z^2 + 9 + 16}{4}} = -\int \frac{dz}{3z^2 + 25}$

Step 4: Evaluate the integral. $f(x) = -\int \frac{dz}{3z^2 + 25} = -\frac{1}{3} \int \frac{dz}{z^2 + \frac{25}{3}} = -\frac{1}{3} \int \frac{dz}{z^2 + \left(\frac{5}{\sqrt{3}}\right)^2}$ Using the integration rule $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$, we have:

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

Step 5: Substitute back for $z$ and $t$. Since $z = \sqrt{4t^2 - 3}$ and $t = \frac{1}{x}$, we have:

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

Step 6: Use the initial condition $f(0) = 0$ to find the constant $C$. Since the limit as $x$ approaches 0 of $\frac{\sqrt{4-3x^2}}{x}$ approaches infinity, we have:

$f(0) = -\frac{1}{5\sqrt{3}} \tan^{-1}(\infty) + C = -\frac{1}{5\sqrt{3}} \cdot \frac{\pi}{2} + C = 0$ Thus, $C = \frac{\pi}{10\sqrt{3}}$.

Step 7: Write the expression for $f(x)$. $f(x) = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{\frac{4-3x^2}{x^2}}\right) + \frac{\pi}{10\sqrt{3}}$

Step 8: Evaluate $f(1)$. $f(1) = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{\frac{4-3}{1}}\right) + \frac{\pi}{10\sqrt{3}} = -\frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}}{5}\right) + \frac{\pi}{10\sqrt{3}}$

Step 9: Use the identity $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$. We can rewrite the expression as:

$f(1) = \frac{1}{5\sqrt{3}} \left(\frac{\pi}{2} - \tan^{-1}\left(\frac{\sqrt{3}}{5}\right)\right) - \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{\sqrt{3}}{5}\right) + \frac{\pi}{10\sqrt{3}} = \frac{1}{5\sqrt{3}} \cot^{-1}\left(\frac{\sqrt{3}}{5}\right) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5}{\sqrt{3}}\right)$

Step 10: Determine $\alpha$ and $\beta$. Comparing $f(1) = \frac{1}{5\sqrt{3}} \tan^{-1}\left(\frac{5}{\sqrt{3}}\right)$ with $f(1) = \frac{1}{\alpha \beta} \tan^{-1}\left(\frac{\alpha}{\beta}\right)$, we have $\alpha = 5$ and $\beta = \sqrt{3}$.

Step 11: Calculate $\alpha^2 + \beta^2$. $\alpha^2 + \beta^2 = 5^2 + (\sqrt{3})^2 = 25 + 3 = 28$.

Common Mistakes & Tips

• Sign Errors: Be careful with the signs when performing substitutions and integrating.
• Simplifying Expressions: Always simplify expressions to avoid unnecessary complexity.
• Using Trigonometric Identities: Remember to use trigonometric identities to manipulate the expressions into the desired form.

Summary

We solved the indefinite integral using a series of substitutions and trigonometric identities. First, we substituted $x = \frac{1}{t}$ to simplify the expression, then $z^2 = 4t^2-3$ to eliminate the square root in the denominator. After integrating with respect to $z$, we substituted back to obtain the expression in terms of $x$. Using the initial condition $f(0)=0$, we determined the constant of integration. Finally, we evaluated $f(1)$ and compared it to the given form to find $\alpha$ and $\beta$, and calculated $\alpha^2 + \beta^2$.

The final answer is \boxed{28}.