1. **State the problem:** Evaluate the integral $$\int \frac{dx}{\sqrt{49x^2 - 81}}$$ with the condition $$x > \frac{9}{7}$$. 2. **Identify the form:** The integral is of the form $$\int \frac{dx}{\sqrt{a^2 x^2 - b^2}}$$ where $$a=7$$ and $$b=9$$. 3. **Recall the formula:** For $$\int \frac{dx}{\sqrt{a^2 x^2 - b^2}}$$ with $$x > \frac{b}{a}$$, the solution is $$\frac{1}{a} \ln \left| a x + \sqrt{a^2 x^2 - b^2} \right| + C$$. 4. **Apply the formula:** Substitute $$a=7$$ and $$b=9$$: $$\int \frac{dx}{\sqrt{49x^2 - 81}} = \frac{1}{7} \ln \left| 7x + \sqrt{49x^2 - 81} \right| + C$$. 5. **Explain the condition:** Since $$x > \frac{9}{7}$$, the expression inside the square root is positive, and the logarithm argument is positive, so absolute value can be omitted. 6. **Final answer:** $$\boxed{\int \frac{dx}{\sqrt{49x^2 - 81}} = \frac{1}{7} \ln \left(7x + \sqrt{49x^2 - 81} \right) + C}$$