Answer
$$ - \frac{{137}}{{20}}$$
Work Step by Step
$$\eqalign{
& \int_1^8 {\frac{{\left( {{x^{1/3}} + 1} \right)\left( {2 - {x^{2/3}}} \right)}}{{{x^{1/3}}}}} dx \cr
& {\text{use distributive property}} \cr
& \int_1^8 {\frac{{2{x^{1/3}} - x + 2 - {x^{2/3}}}}{{{x^{1/3}}}}} dx \cr
& \int_1^8 {\left( {2 - {x^{2/3}} + 2{x^{ - 1/3}} - {x^{1/3}}} \right)} dx \cr
& {\text{integrate by using }}\int_a^b {{x^n}} dx = \left( {\frac{{{x^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \left( {2x - \frac{{3{x^{5/3}}}}{5} + 3{x^{2/3}} - \frac{3}{4}{x^{4/3}}} \right)_1^8 \cr
& {\text{Evaluating, we get:}} \cr
& = \left( {2\left( 8 \right) - \frac{{3{{\left( 8 \right)}^{5/3}}}}{5} + 3{{\left( 8 \right)}^{2/3}} - \frac{3}{4}{{\left( 8 \right)}^{4/3}}} \right) - \left( {2\left( 1 \right) - \frac{{3{{\left( 1 \right)}^{5/3}}}}{5} + 3{{\left( 1 \right)}^{2/3}} - \frac{3}{4}{{\left( 1 \right)}^{4/3}}} \right) \cr
& = - \frac{{16}}{5} - \frac{{73}}{{20}} \cr
& = - \frac{{137}}{{20}} \cr} $$
