11. $int\frac{dx}{xsqrt{x^{2}-9}}$

Explanation:

Step1: Use substitution

Let $x = 3\sec\theta$, then $dx=3\sec\theta\tan\theta d\theta$. And $\sqrt{x^{2}-9}=\sqrt{9\sec^{2}\theta - 9}=3\tan\theta$. The integral $\int\frac{dx}{x\sqrt{x^{2}-9}}$ becomes $\int\frac{3\sec\theta\tan\theta d\theta}{3\sec\theta\cdot3\tan\theta}$.

Step2: Simplify the integrand

$\int\frac{3\sec\theta\tan\theta d\theta}{3\sec\theta\cdot3\tan\theta}=\int\frac{1}{3}d\theta$.

Step3: Integrate

$\int\frac{1}{3}d\theta=\frac{1}{3}\theta + C$.

Step4: Back - substitute

Since $x = 3\sec\theta$, then $\sec\theta=\frac{x}{3}$ and $\theta=\text{arcsec}(\frac{x}{3})$. So the integral is $\frac{1}{3}\text{arcsec}(\frac{x}{3})+C$.

Answer:

$\frac{1}{3}\text{arcsec}(\frac{x}{3})+C$