Question

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

Answer 1

Answer: 5

$f(x)={\log }_2\left(x^3+1\right)=y$
$x^3+1=2^y\Rightarrow x={\left(2^y-1\right)}^{1/3}=f^{-1}(y)$
$f^{-1}(x)={\left(2^x-1\right)}^{1/3}$
$={\int }_1^2{\log }_2\left(x^3+1\right)dx+\underset{1}{\overset{{\log }_{2}9}{\int }}{\left(2^x-1\right)}^{1/3}dx$
$={\int }_1^{2}f(x)dx+\underset{1}{\overset{{\log }_{2}9}{\int }}{f}^{-1}(x)dx=2{\log }_{2}9-1$
$=8<9<2^{7/2}\Rightarrow 3<{\log }_{2}9<\frac{7}{2}$
$=5<2{\log }_{2}9-1<6$
$\left[2{\log }_{2}9-1\right]=5$