evaluate ∫ (√(x² - 49) / x) dx.

# Explanation: ## Step1: Use trig - substitution Let $x = 7\sec\theta$, then $dx=7\sec\theta\tan\theta d\theta$. And $\sqrt{x^{2}-49}=\sqrt{49\sec^{2}\theta - 49}=7\tan\theta$. ## Step2: Rewrite the integral The integral $\int\frac{\sqrt{x^{2}-49}}{x}dx$ becomes $\int\frac{7\tan\theta}{7\sec\theta}\cdot7\sec\theta\tan\theta d\theta=\int7\tan^{2}\theta d\theta$. ## Step3: Use the identity $\tan^{2}\theta=\sec^{2}\theta - 1$ $\int7\tan^{2}\theta d\theta=7\int(\sec^{2}\theta - 1)d\theta=7(\tan\theta-\theta)+C$. ## Step4: Back - substitute Since $x = 7\sec\theta$, then $\sec\theta=\frac{x}{7}$ and $\theta=\text{arcsec}(\frac{x}{7})$. Also, $\tan\theta=\frac{\sqrt{x^{2}-49}}{7}$. So the integral is $\sqrt{x^{2}-49}-7\text{arcsec}(\frac{x}{7})+C$. # Answer: $\sqrt{x^{2}-49}-7\text{arcsec}(\frac{x}{7})+C$