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:

$\sqrt{x^{2}-49}-7\text{arcsec}(\frac{x}{7})+C$