Let the domain of the function $f(x)=\log _2 \log _4 \log _6\left(3+4 x-x^2\right)$ be $(a, b)$. If $\int_0^{b-a}\left[x^2\right] d x=p-\sqrt{q}-\sqrt{r}, p, q, r \in \mathbb{N}, \operatorname{gcd}(p, q, r)=1$, where $[\cdot]$ is the greatest integer function, then $p+q+r$ is equal to

<p><strong>Step 1</strong>: Ensure the innermost function is greater than zero: </p> <p>$ \log_6(3 + 4x - x^2) > 1 $</p></p> <p><p><strong>Step 2</strong>: Simplify the inequality from Step 1:</p> <p>$ 3 + 4x - x^2 > 6 \implies x^2 - 4x + 3 < 0 $</p></p> <p><p><strong>Step 3</strong>: Solve the quadratic inequality:</p> <p>$ (x - 1)(x - 3) < 0 $</p> <p>This inequality indicates that $ x $ must lie between the roots, giving the interval $ x \in (1, 3) $.</p></p> <p>With the domain of $ f(x) $ identified as $ (a, b) = (1, 3) $, we calculate the definite integral over $[0, b-a]$:</p> <h3>Calculate the Integral:</h3> <p>Given:</p> <p>$ \int_0^{b-a} [x^2] \, dx $</p> <p>For $ b - a = 2 $, we compute:</p> <p>$ \int_1^2 [x^2] \, dx = \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^2 3 \, dx $</p> <h3>Computation of Each Integral Segment:</h3> <p><p>$\int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1$</p></p> <p><p>$\int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2(\sqrt{3} - \sqrt{2})$</p></p> <p><p>$\int_{\sqrt{3}}^2 3 \, dx = 3(2 - \sqrt{3})$</p></p> <p>Summing these, we have:</p> <p>$ 5 - \sqrt{2} - \sqrt{3} $</p> <h3>Conclusion:</h3> <p>The values for $ p, q, $ and $ r $ are $ p = 5 $, $ q = 2 $, and $ r = 3 $, with the greatest common divisor of these numbers being 1. Therefore, adding them together gives:</p> <p>$ p + q + r = 5 + 2 + 3 = 10 $</p>