1. The problem is to evaluate the definite integral $$\int_4^6 \sqrt{6 + x^2} \, dx$$. 2. We use the formula for the integral of $$\sqrt{a^2 + x^2}$$: $$\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln\left|x + \sqrt{x^2 + a^2}\right| + C$$ 3. Here, $$a^2 = 6$$, so $$a = \sqrt{6}$$. 4. Applying the formula, the integral becomes: $$\int_4^6 \sqrt{6 + x^2} \, dx = \left[ \frac{x}{2} \sqrt{x^2 + 6} + \frac{6}{2} \ln\left|x + \sqrt{x^2 + 6}\right| \right]_4^6$$ 5. Calculate the expression at the upper limit $$x=6$$: $$\frac{6}{2} \sqrt{6^2 + 6} + 3 \ln\left(6 + \sqrt{36 + 6}\right) = 3 \sqrt{42} + 3 \ln(6 + \sqrt{42})$$ 6. Calculate the expression at the lower limit $$x=4$$: $$\frac{4}{2} \sqrt{4^2 + 6} + 3 \ln\left(4 + \sqrt{16 + 6}\right) = 2 \sqrt{22} + 3 \ln(4 + \sqrt{22})$$ 7. Subtract the lower limit value from the upper limit value: $$3 \sqrt{42} + 3 \ln(6 + \sqrt{42}) - \left(2 \sqrt{22} + 3 \ln(4 + \sqrt{22})\right)$$ 8. This is the exact value of the definite integral. Final answer: $$\int_4^6 \sqrt{6 + x^2} \, dx = 3 \sqrt{42} + 3 \ln(6 + \sqrt{42}) - 2 \sqrt{22} - 3 \ln(4 + \sqrt{22})$$