The greatest integer less than or equal to ${\int }_1^2{\log }_2\left(x^3+1\right)dx+{\int }_1^{{\log }_{2}9}{\left(2^x-1\right)}^{\frac{1}{3}}dx$ is

Let $I={\int }_1^2{\log }_2\left(x^3+1\right)dx+{\int }_1^{{\log }_{2}9}{\left(2^x-1\right)}^{\frac{1}{3}}dx$

Let ${\int }_1^{{\log }_{2}9}{\left(2^x-1\right)}^{\frac{1}{3}}dx=I_1$

Let $2^x-1=t^3$

$\Rightarrow 2^x\ln 2dx=3t^{2}dt$

$\Rightarrow dx=\frac{3t^2}{\left(t^3+1\right)\ln 2}dt$

i.e. $I_1={\int }_1^2\frac{3t^3}{\ln 2\left(t^3+1\right)}dt$ 

So, $I={\int }_1^2{\log }_2\left(x^3+1\right)dx+{\int }_1^2\frac{3t^3}{\ln 2\left(t^3+1\right)}dt$

or $I={\int }_1^2\left({\log }_2\left(t^3+1\right)+t·\frac{3t^2}{\left(t^3+1\right)\ln 2}\right)dt$

$={\left[t\cdot {\log }_2\left(t^3+1\right)\right]}_1^2$

$=2{\log }_{2}9-1{\log }_{2}2$

$=2{\log }_{2}9-1=4{\log }_{2}3-1$

So, $\left[I\right]=5$