The greatest integer less than or equal to $$\displaystyle\int_1^2 \log_2(x^3 + 1) \, dx + \int_1^{\log_2 9} (2^x - 1)^{1/3} \, dx$$ is _______.

Write the first integral with natural logarithm:
$$I_1 = \int_{1}^{2} \log_2\!\left(x^3+1\right)\,dx = \frac{1}{\ln 2}\int_{1}^{2}\ln\!\left(x^3+1\right)\,dx.$$ Call the natural-log integral $$J_1=\displaystyle\int_{1}^{2}\ln\!\left(x^3+1\right)\,dx.$$ Hence $$I_1=\dfrac{J_1}{\ln 2}.$$

For the second integral put $$t=(2^x-1)^{1/3}\qquad\Bigl(\;x=1\to t=1,\;x=\log_29\to t=2\Bigr).$$ Then $$2^x=t^3+1,\qquad dt=\frac{1}{3}(2^x-1)^{-2/3}2^x\ln2\,dx =\frac{(t^3+1)\ln2}{3t^2}\,dx.$$ Thus $$dx=\frac{3t^2}{(t^3+1)\ln2}\,dt.$$ The integral becomes $$I_2=\int_{1}^{\log_29}(2^x-1)^{1/3}\,dx =\frac{3}{\ln2}\int_{1}^{2}\frac{t^3}{t^3+1}\,dt.$$ Define $$J_2=3\int_{1}^{2}\frac{t^3}{t^3+1}\,dt,$$ so that $$I_2=\dfrac{J_2}{\ln2}.$$

Now observe that the required sum is $$I_1+I_2=\frac{J_1+J_2}{\ln2}.$$

Relation between $$J_1$$ and $$J_2$$:
Integrate $$J_1$$ by parts with $$u=\ln(1+t^3),\;dv=dt.$$ $$$ J_1=\Bigl[t\ln(1+t^3)\Bigr]_{1}^{2}-\int_{1}^{2}\frac{3t^3}{1+t^3}\,dt =\Bigl[t\ln(1+t^3)\Bigr]_{1}^{2}-J_2. $$$ Hence $$J_1+J_2=\Bigl[t\ln(1+t^3)\Bigr]_{1}^{2}.$$ Evaluate the boundary term: $$$ t=2:\;2\ln(1+8)=2\ln9,\qquad t=1:\;1\ln(1+1)=\ln2. $$$ Therefore $$J_1+J_2=2\ln9-\ln2=\ln\!\left(\frac{9^2}{2}\right)=\ln(40.5).$$

Finally, $$I_1+I_2=\frac{\ln(40.5)}{\ln2}=\log_{2}(40.5).$$ Because $$2^5=32 \lt 40.5 \lt 64=2^6,$$ we have $$5 \lt \log_{2}(40.5) \lt 6.$$

The greatest integer less than or equal to the given expression is
5.