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

1. Find the indefinite integral using substitution.

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

$\displaystyle\int \frac{(\tan^{-1}2x)^2}{1+4x^2}\,dx = \frac{1}{2} \int (\tan^{-1}2x)^2 \left( \frac{2}{1+4x^2} \right) \,dx$
$\displaystyle= \frac{1}{2} \int u^2 \,du$
$\displaystyle= \frac{1}{6}u^3 + C$
$\displaystyle= \frac{1}{6}(\tan^{-1}2x)^3 + C$

2. Evaluate the definite integral to obtain the formula in terms of $n$.

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

3. Evaluate the integral to infinity by taking the limit as $n \to \infty$.

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

$\displaystyle\int_{\tfrac{1}{2}}^{\infty} \frac{(\tan^{-1}2x)^2}{1+4x^2}\,dx = \lim_{n \to \infty} \left( \frac{1}{6}(\tan^{-1}2n)^3 - \frac{\pi^3}{384} \right)$
$= \frac{1}{6}\left(\frac{\pi}{2}\right)^3 - \frac{\pi^3}{384}$
$= \frac{\pi^3}{48} - \frac{\pi^3}{384}$
$= \frac{8\pi^3 - \pi^3}{384}$
$= \frac{7\pi^3}{384}$