Obtain a formula for $\int_{\frac{1}{2}}^{n} \frac{(\tan^{-1} 2x)^2}{1 + 4x^2} \,\mathrm{d}x$ in terms of $n$, where $n > 0$. Hence evaluate $\int_{\frac{1}{2}}^{\infty} \frac{(\tan^{-1} 2x)^2}{1 + 4x^2} \,\mathrm{d}x$ exactly.

1. Apply integration by substitution to find the definite integral in terms of $n$.

Let $u = \tan^{-1} 2x$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{1 + (2x)^2} = \frac{2}{1 + 4x^2} \implies \frac{1}{1 + 4x^2}\,\mathrm{d}x = \frac{1}{2}\,\mathrm{d}u$.

Change of limits:
When $x = \frac{1}{2}$, $u = \tan^{-1}(1) = \frac{\pi}{4}$.
When $x = n$, $u = \tan^{-1}(2n)$.

$\int_{\frac{1}{2}}^{n} \frac{(\tan^{-1} 2x)^2}{1 + 4x^2} \,\mathrm{d}x = \int_{\frac{\pi}{4}}^{\tan^{-1} 2n} u^2 \left( \frac{1}{2} \right) \mathrm{d}u$
$= \left[ \frac{u^3}{6} \right]_{\frac{\pi}{4}}^{\tan^{-1} 2n}$
$= \frac{1}{6} (\tan^{-1} 2n)^3 - \frac{1}{6}\left(\frac{\pi}{4}\right)^3$
$= \frac{1}{6} (\tan^{-1} 2n)^3 - \frac{\pi^3}{384}$

2. Evaluate the improper integral by taking the limit as $n \to \infty$.

$\int_{\frac{1}{2}}^{\infty} \frac{(\tan^{-1} 2x)^2}{1 + 4x^2} \,\mathrm{d}x = \lim_{n \to \infty} \left( \frac{1}{6} (\tan^{-1} 2n)^3 - \frac{\pi^3}{384} \right)$

As $n \to \infty$, $2n \to \infty$, so $\tan^{-1} 2n \to \frac{\pi}{2}$.

$\lim_{n \to \infty} \left( \frac{1}{6} (\tan^{-1} 2n)^3 - \frac{\pi^3}{384} \right) = \frac{1}{6}\left(\frac{\pi}{2}\right)^3 - \frac{\pi^3}{384}$
$= \frac{\pi^3}{48} - \frac{\pi^3}{384}$
$= \frac{8\pi^3 - \pi^3}{384} = \frac{7\pi^3}{384}$