5. evaluate the integral ∫ dx / √(x² - 9).

Explanation:

Step1: Use trig - substitution

Let $x = 3\sec\theta$, then $dx=3\sec\theta\tan\theta d\theta$. Also, $\sqrt{x^{2}-9}=\sqrt{9\sec^{2}\theta - 9}=3\tan\theta$ (since $\sec^{2}\theta-1 = \tan^{2}\theta$ and for the appropriate domain of $\theta$).

Step2: Substitute into the integral

The integral $\int\frac{dx}{\sqrt{x^{2}-9}}$ becomes $\int\frac{3\sec\theta\tan\theta d\theta}{3\tan\theta}=\int\sec\theta d\theta$.

Step3: Integrate $\sec\theta$

The antiderivative of $\sec\theta$ is $\ln|\sec\theta+\tan\theta| + C$.

Step4: Back - substitute

Since $x = 3\sec\theta$, then $\sec\theta=\frac{x}{3}$ and $\tan\theta=\frac{\sqrt{x^{2}-9}}{3}$. So $\ln|\sec\theta+\tan\theta|+C=\ln\left|\frac{x}{3}+\frac{\sqrt{x^{2}-9}}{3}\right|+C=\ln|x + \sqrt{x^{2}-9}|-\ln(3)+C$. Since $-\ln(3)+C$ is still a constant, the final result is $\ln|x+\sqrt{x^{2}-9}|+C$.

Answer:

$\ln|x+\sqrt{x^{2}-9}|+C$