TITLE: Evaluate $\int_2^4\frac{\sqrt{x^2-4}}{x^2}\mathrm dx$
QUESTION [5 upvotes]: Evaluate $$\int\limits_2^4\frac{\sqrt{x^2-4}}{x^2}\mathrm dx$$
My working:
$x=2\sec\theta\quad\Rightarrow\quad\theta=\arccos\left(\frac{2}{x}\right)$
$dx=2\sec\theta\tan\theta d\theta$
$I=\int\frac{\sqrt{4\sec^2\theta-4}}{4\sec^2\theta}2\sec\theta\tan\theta d\theta=\int\frac{\tan^2\theta}{\sec\theta}d\theta=\int\frac{\sin^2\theta}{\cos\theta}d\theta=\int\sec\theta d\theta-\int\cos\theta d\theta\\=\ln|\sec\theta+\tan\theta|-\sin\theta+C\\=\left.\ln\left|\frac{x}{2}+\frac{\sqrt{1-(2/x)^2}}{2/x}\right|-\sqrt{1-\left(\frac{2}{x}\right)^2}\right]_2^4$

EDIT
$=\ln\left|\frac{4+\sqrt{12}}{2}\right|-\sqrt{1-\frac{1}{4}}-\ln\left|\frac{2+\sqrt{0}}{2}\right|-\sqrt{1-1}\\=\ln|2+\sqrt{3}|-\frac{\sqrt{3}}{2}\qquad\blacksquare$

REPLY [2 votes]: Yes, you are completely right, except the $-1$ term. You have:
\begin{align}
& \left.\ln\left|\frac{x}{2}+\frac{\sqrt{1-(2/x)^2}}{2/x}\right|-\sqrt{1-\left(\frac{2}{x}\right)^2}\right|_2^4 = \\
= & \ln\left|\frac{4+\sqrt{12}}{2}\right|-\sqrt{1-\frac{1}{4}}-\ln\left|\frac{2+\sqrt{0}}{2}\right|-\color{red}{\sqrt{1-0}}
\end{align}
You divide $2$ by $2$ and get's $0$ under the square root sign, it should be $\sqrt{1-1}$, and the result:
$$
\ln|2+\sqrt{3}|-\frac{\sqrt{3}}{2}\blacksquare
$$