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

1. Set up the substitution and find the differential.

Let $u = \tan^{-1} 2x$.
Then $\frac{du}{dx} = \frac{2}{1+(2x)^2} = \frac{2}{1+4x^2}$.
Rearranging gives $dx = \frac{1+4x^2}{2} \, du$.

2. Determine the new limits of integration corresponding to $u$.

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

3. Integrate and substitute the new limits to obtain the formula in terms of $n$.

$\displaystyle\int_{\frac{1}{2}}^{n} \frac{(\tan^{-1} 2x)^2}{1+4x^2} \, dx = \int_{\frac{\pi}{4}}^{\tan^{-1}(2n)} \frac{u^2}{1+4x^2} \left( \frac{1+4x^2}{2} \right) \, du$

$= \frac{1}{2} \displaystyle\int_{\frac{\pi}{4}}^{\tan^{-1}(2n)} u^2 \, du$

$= \frac{1}{2} \left[ \frac{u^3}{3} \right]_{\frac{\pi}{4}}^{\tan^{-1}(2n)}$

$= \frac{1}{6} (\tan^{-1} 2n)^3 - \frac{1}{6} \left(\frac{\pi}{4}\right)^3$

$= \frac{1}{6} (\tan^{-1} 2n)^3 - \frac{\pi^3}{384}$

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

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

As $n \to \infty$, $2n \to \infty$, which implies $\tan^{-1}(2n) \to \frac{\pi}{2}$.

$= \frac{1}{6} \left(\frac{\pi}{2}\right)^3 - \frac{\pi^3}{384}$

$= \frac{\pi^3}{48} - \frac{\pi^3}{384}$

$= \frac{8\pi^3 - \pi^3}{384} = \frac{7\pi^3}{384}$