Question: Solve with solution
1. **State the problem:** Evaluate the definite integral $$\int_3^6 \frac{x}{3\sqrt{x^2 - 8}} \, dx$$.
2. **Recall the formula and rules:** The integral involves a radical in the denominator. We can simplify the integral by substitution or algebraic manipulation.
3. **Rewrite the integral:**
$$\int_3^6 \frac{x}{3\sqrt{x^2 - 8}} \, dx = \frac{1}{3} \int_3^6 \frac{x}{\sqrt{x^2 - 8}} \, dx$$
4. **Use substitution:** Let $$u = x^2 - 8$$, then $$du = 2x \, dx$$, so $$x \, dx = \frac{du}{2}$$.
5. **Change the limits:** When $$x=3$$, $$u = 3^2 - 8 = 9 - 8 = 1$$.
When $$x=6$$, $$u = 6^2 - 8 = 36 - 8 = 28$$.
6. **Rewrite the integral in terms of $$u$$:**
$$\frac{1}{3} \int_{u=1}^{28} \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{6} \int_1^{28} u^{-1/2} \, du$$
7. **Integrate:**
$$\int u^{-1/2} \, du = 2 u^{1/2} + C$$
8. **Apply the definite integral:**
$$\frac{1}{6} \times 2 \left[ u^{1/2} \right]_1^{28} = \frac{1}{3} \left( \sqrt{28} - \sqrt{1} \right) = \frac{1}{3} (2\sqrt{7} - 1)$$
9. **Final answer:**
$$\boxed{\frac{2\sqrt{7} - 1}{3}}$$