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}}d\mathrm{x. }

is _______________ .

The correct answer is 5.

Let us evaluate the given expression step-by-step.
We are asked to find the greatest integer less than or equal to:

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

Let us analyze the function $f\left(x\right)={log}_2(x^3+1)$.
Let $y=f\left(x\right)={log}_2(x^3+1)$.
To find the inverse function $f^{-1}\left(x\right)$, solve for $x$ in terms of $y$:

$2^y=x^3+1$

$x^3=2^y-1$

$x={(2^y-1)}^{\frac{1}{3}}$

Thus, the inverse function is $f^{-1}\left(x\right)={(2^x-1)}^{\frac{1}{3}}$.

Now, check the limits of integration:
When $x=1$, $f\left(1\right)={log}_2(1^3+1)={log}_{2}2=1$.
When $x=2$, $f\left(2\right)={log}_2(2^3+1)={log}_{2}9$.

Using the fundamental identity for definite integrals involving inverse functions:

${\int }_a^{b}f\left(x\right)dx+{\int }_{f\left(a\right)}^{f\left(b\right)}{f}^{-1}\left(x\right)dx=b·f\left(b\right)-a·f\left(a\right)$

Substituting $a=1$ and $b=2$:

$I=2·f\left(2\right)-1·f\left(1\right)$

$I=2·{log}_{2}9-1·1$

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

Now we calculate the numeric value of $I$ to find its greatest integer function value:
Since $2^3=8$ and $2^{3.17}\approx 9$, ${log}_{2}9\approx 3.17$.

$I=2(3.17)-1=6.34-1=5.34$

The greatest integer less than or equal to $I$ is:

$\left[I\right]=[5.34]=5$