Let $$I = \int_1^2 {{{\log }_2}({x^3} + 1)dx + \int_1^{\log _2^9} {{{({2^x} - 1)}^{{1 \over 3}}}dx} } $$

Let $${I_1} = \int_1^2 {{{\log }_2}({x^3} + 1)dx} $$

and $${I_2} = \int_1^{\log _2^9} {{{({2^x} - 1)}^{{1 \over 3}}}dx} $$

Let $${2^x} - 1 = {y^3}$$

$$ \Rightarrow {2^x}\,.\,\log _2^2 = 3{y^2}\,.\,{{dy} \over {dx}}$$

$$ \Rightarrow dx = {{3{y^2}} \over {{2^x}}}dy$$

$$ \Rightarrow dx = {{3{y^2}dy} \over {{y^3} + 1}}$$

When $$x = 1$$ then $${y^3} = {2^1} - 1 = 1 \Rightarrow y = 1$$

When $$x = \log _2^9$$ then $${y^3} = {2^{\log _2^9}} - 1 \Rightarrow {y^3} = 8 \Rightarrow y = 2$$

$$\therefore$$ $${I_2} = \int_1^2 {y\,.\,{{3{y^2}} \over {{y^3} + 1}}dy} $$

$$ = \int_1^2 {{{3{y^3}dy} \over {{y^3} + 1}}} $$

Let $$y = x$$

$$ \Rightarrow dy = dx$$

$$\therefore$$ $${I_2} = \int_1^2 {{{3{x^3}dx} \over {{x^3} + 1}}} $$

Now, $${I_1} = \int_1^2 {{{\log }_2}({x^3} + 1)dx} $$

$$ = \left[ {{{\log }_2}({x^3} + 1)\,.\,x} \right]_1^2 - \int_1^2 {{1 \over {{x^3} + 1}}\,.\,{{3{x^2}} \over {\log _2^2}}\,.\,x\,dx} $$

$$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_2^2 - \int_1^2 {{{3{x^3}dx} \over {{x^3} + 1}}} $$

$$\therefore$$ $$I = {I_1} + {I_2}$$

$$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_1^2 - \int_1^2 {{{3{x^3}dx} \over {({x^3} + 1)}} + \int_1^2 {{{3{x^3}dx} \over {({x^3} + 1)}}} } $$

$$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_1^2$$

$$ = 2\log _2^9 - 1\,.\,\log _2^2$$

$$ = 2\log _2^9 - 1$$

$$ = 4\log _2^3 - 1$$

$$ = 4 \times 1.58 - 1$$

$$ = 6.32 - 1$$

$$ = 5.32$$

$$\therefore$$ Greatest integer value of

$$I = [5.32] = 5$$

Other Method :-

Let $$f(x) = {\log _2}({x^3} + 1) = y$$

$$ \Rightarrow {x^3} + 1 = {2^y}$$

$$ \Rightarrow {x^3} = {2^y} - 1$$

$$ \Rightarrow x = {\left( {{2^y} - 1} \right)^{{1 \over 3}}}$$

$$\therefore$$ $${f^{ - 1}}(x) = {\left( {{2^x} - 1} \right)^{{1 \over 3}}}$$

And $$f(1) = \log _2^{(1 + 1)} = 1 = f(a)$$ (Assume)

$$f(2) = \log _2^{(8 + 1)} = \log _2^9 = f(b)$$

$$\therefore$$ $$I = \int_a^b {f(x)dx + \int_{f(a)}^{f(b)} {{f^{ - 1}}(x)dx} } $$

Let $${f^{ - 1}}(x) = t$$

$$ \Rightarrow x = f(t)$$

$$ \Rightarrow dx = f'(t)dt$$

When $$x = f(a)$$ then $$f(a) = f(t) \Rightarrow t = a$$

When $$x = f(b)$$ then $$f(b) = f(t) \Rightarrow t = b$$

$$\therefore$$ $$I = \int_a^b {f(t)dt + \int_a^b {t\,.\,f'(t)dt} } $$

$$ = \int_a^b {\left( {f(t) + t\,.\,f'(t)} \right)dt} $$

$$ = \left[ {t\,.\,f(t)} \right]_a^b$$

$$ = b\,.\,f(b) - a\,.\,f(a)$$

Here $$b = 2$$, $$f(b) = \log _2^9$$

and $$a = 1$$, $$f(a) = 1$$

$$\therefore$$ $$I = 2\log _2^9 - 1\,.\,1$$

$$ = 4\,.\,\log _2^3 - 1$$

$$ = 4 \times 1.58 - 1$$

$$ = 6.32 - 1$$

$$ = 5.32$$

$$\therefore$$ $$[I] = 5$$