The greatest integer less than or equal to $\int_{1}^{2}\log_{2}(x^{3}+1)\,dx+\int_{1}^{\log_{2}9}(2^{x}-1)^{\frac{1}{3}}\,dx$ is ________.

The greatest integer less than or equal to ∫₁²log₂(x³+1) dx+∫₁^(log₂9)(2ˣ-1)^(1/3) dx is ________. $\text{Given: } I = \displaystyle\int_{1}^{2}\log_{2}(x^{3}+1)\,dx + \int_{1}^{\log_{2}9}(2^{x}-1)^{1/3}\,dx$ $\text{To find: } \left\lfloor I \right\rfloor$ $1.\ \text{the integrands are inverses}$ $2.\ \text{inverse-function area identity}$ $3.\ \text{bound the value} \Rightarrow \lfloor I \rfloor$ $y = \log_2(x^3+1) \iff x = (2^y - 1)^{1/3}$ $\text{so the two integrands are inverse functions}$ $f(1) = 1,\quad f(2) = \log_2 9;\qquad f^{-1}(1) = 1,\quad f^{-1}(\log_2 9) = 2$ $\int_a^b f(x)\,dx + \int_{f(a)}^{f(b)} f^{-1}(y)\,dy = b\,f(b) - a\,f(a)$ $I = 2\log_2 9 - 1$ $2^3 = 8 < 9 < 16 = 2^4 \Rightarrow 3 < \log_2 9 < 4$ $2^{3.5} = 8\sqrt2 \approx 11.3 > 9 \Rightarrow 3 < \log_2 9 < 3.5$ $\Rightarrow 5 < I < 6$ $\left\lfloor I \right\rfloor = 5$ $\boxed{5}$