Question

$\int\limits_2^4 \frac{\sqrt{x^2-4}}{x} d x$ equal to

• A. $2(3 \sqrt{3}-\pi)$
• B. $2 \sqrt{3}-\pi$
• C. $ \frac{2}{3}(3 \sqrt{3}-\pi)$
• D. $\pi$

Answer 1

Answer: (C)

Put $x=2 \sec \theta$
$\Rightarrow \int\limits_2^4 \frac{\sqrt{x^2-4}}{x} d x =\int\limits_0^{\pi / 3} \frac{2 \tan \theta}{2 \sec \theta} \cdot 2 \sec \theta \tan \theta d \theta$
$=\int\limits_0^{\pi / 3} 2 \tan ^2 \theta d \theta=2 \int\limits_0^{\pi / 3}\left(\sec ^2 \theta-1\right) d \theta $
$=2(\tan \theta-\theta) \|_0^{\pi / 3}=2\left(\left(\sqrt{3}-\frac{\pi}{3}\right)-(0-0)\right)=\frac{2}{3}(3 \sqrt{3}-\pi)$