evaluate the integral.\n int\frac{dx}{x^{2}sqrt{x^{2}-4}},x > 2\n int\frac{dx}{x^{2}sqrt{x^{2}-4}}=square

Explanation:

Step1: Use trigonometric substitution

Let $x = 2\sec\theta$, then $dx=2\sec\theta\tan\theta d\theta$. Also, $\sqrt{x^{2}-4}=\sqrt{4\sec^{2}\theta - 4}=2\tan\theta$ and $x^{2}=4\sec^{2}\theta$.

Step2: Substitute into the integral

The integral $\int\frac{dx}{x^{2}\sqrt{x^{2}-4}}$ becomes $\int\frac{2\sec\theta\tan\theta d\theta}{4\sec^{2}\theta\cdot2\tan\theta}=\frac{1}{4}\int\frac{d\theta}{\sec\theta}=\frac{1}{4}\int\cos\theta d\theta$.

Step3: Integrate $\cos\theta$

$\frac{1}{4}\int\cos\theta d\theta=\frac{1}{4}\sin\theta + C$.

Step4: Back - substitute $\theta$ in terms of $x$

Since $x = 2\sec\theta$, then $\sec\theta=\frac{x}{2}$ and $\cos\theta=\frac{2}{x}$. Using $\sin^{2}\theta=1 - \cos^{2}\theta$, we have $\sin\theta=\frac{\sqrt{x^{2}-4}}{x}$. So the integral is $\frac{\sqrt{x^{2}-4}}{4x}+C$.

Answer:

$\frac{\sqrt{x^{2}-4}}{4x}+C$