Question

The greatest integer less than or equal to
$\int\limits_1^2 \log _2\left(x^3+1\right) d x+\int\limits_1^{\log _2 9}\left(2^x-1\right)^{\frac{1}{3}} dx$
is _____

Answer 1

$ 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 +\int\limits_1^{\log _2 9}\left(2^x-1\right)^{1 / 3} dx$
$ =\int_1^2 f ( x ) dx +\int\limits_1^{\log _2 9} 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}$