We are asked to evaluate the definite integral $\int_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}} \frac{1}{4x^2 + 49} dx$.

1. Problem Description

We are asked to evaluate the definite integral $\int_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}} \frac{1}{4x^2 + 49} dx$.

2. Solution Steps

First, rewrite the integral as follows:

$\int_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}} \frac{1}{4x^2 + 49} dx = \int_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}} \frac{1}{4(x^2 + \frac{49}{4})} dx = \frac{1}{4} \int_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}} \frac{1}{x^2 + (\frac{7}{2})^2} dx$

We know that $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$. Here, $a = \frac{7}{2}$.

Therefore,

$\frac{1}{4} \int_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}} \frac{1}{x^2 + (\frac{7}{2})^2} dx = \frac{1}{4} \left[ \frac{1}{\frac{7}{2}} \arctan(\frac{x}{\frac{7}{2}}) \right]_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}} = \frac{1}{4} \left[ \frac{2}{7} \arctan(\frac{2x}{7}) \right]_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}} = \frac{1}{14} \left[ \arctan(\frac{2x}{7}) \right]_{\frac{7}{2\sqrt{3}}}^{\frac{7\sqrt{3}}{2}}$

Now, evaluate the expression at the upper and lower limits:

$\frac{1}{14} \left[ \arctan(\frac{2(\frac{7\sqrt{3}}{2})}{7}) - \arctan(\frac{2(\frac{7}{2\sqrt{3}})}{7}) \right] = \frac{1}{14} \left[ \arctan(\frac{7\sqrt{3}}{7}) - \arctan(\frac{7}{7\sqrt{3}}) \right] = \frac{1}{14} \left[ \arctan(\sqrt{3}) - \arctan(\frac{1}{\sqrt{3}}) \right]$

We know that $\arctan(\sqrt{3}) = \frac{\pi}{3}$ and $\arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.

Therefore, $\frac{1}{14} \left[ \frac{\pi}{3} - \frac{\pi}{6} \right] = \frac{1}{14} \left[ \frac{2\pi - \pi}{6} \right] = \frac{1}{14} \left[ \frac{\pi}{6} \right] = \frac{\pi}{84}$.

3. Final Answer

$\frac{\pi}{84}$