The greatest integer less than or equal to ∫ _ 1 ^ 2 _ 2 ( x ^ 3 +...

Let $I = {∫}_{1}^{2}\log_{2}\left({x}^{3} + 1\right)dx + {∫}_{1}^{\log_{2}9}\left({2}^{x} - 1\right)\dfrac{1}{3}dx$

Let ${∫}_{1}^{\log_{2}9}\left({2}^{x} - 1\right)\dfrac{1}{3}dx = {I}_{1}$

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

$⇒ {2}^{x}\ln 2dx = 3{t}^{2}dt$

$⇒ dx = \dfrac{3{t}^{2}}{\left({t}^{3} + 1\right)\text{ ln }2}$dt$</p><p>i.e.${I}_{1} = {∫}_{1}^{2}\dfrac{3{t}^{3}}{\ln 2\left({t}^{3} + 1\right)}dt$</p><p>So,$I = {∫}_{1}^{2}\log_{2}\left({x}^{3} + 1\right)dx + {∫}_{1}^{2}\dfrac{3{t}^{3}}{\ln 2\left({t}^{3} + 1\right)}dt$</p><p>or$I = {∫}_{1}^{2}\left(\log_{2}\left({t}^{3} + 1\right) + t \cdot \dfrac{3{t}^{2}}{\left({t}^{3} + 1\right)\ln 2}\right)dt$</p><p>$= {\left[t ⋅ \log_{2}\left({t}^{3} + 1\right)\right]}_{1}^{2}$</p><p>$= 2\log_{2}9 - 1\log_{2}2$</p><p>$= 2\log_{2}9 - 1 = 4\log_{2}3 - 1$</p><p>So, \left[I\right] = 5$