Answer
$\frac{1}{{12}}{\left( {{x^2} + \frac{2}{x}} \right)^6} + C$
Work Step by Step
$$\eqalign{
& \int {\left( {x - \frac{1}{{{x^2}}}} \right)} {\left( {{x^2} + \frac{2}{x}} \right)^5}dx \cr
& {\text{Let }}u = {x^2} + \frac{2}{x},{\text{ then }}du = \left( {2x - \frac{2}{{{x^2}}}} \right)dx \cr
& du = 2\left( {x - \frac{1}{{{x^2}}}} \right)dx \cr
& \frac{1}{2}du = \left( {x - \frac{1}{{{x^2}}}} \right)dx \cr
& {\text{Applying the substitution}} \cr
& \int {{{\left( {{x^2} + \frac{2}{x}} \right)}^5}} \left( {x - \frac{1}{{{x^2}}}} \right)dx = \int {{u^5}} \left( {\frac{1}{2}} \right)du \cr
& = \frac{1}{2}\int {{u^5}} du \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& \frac{1}{2}\int {{u^5}} du = \frac{1}{2}\left( {\frac{{{u^6}}}{6}} \right) + C \cr
& = \frac{1}{{12}}{u^6} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = {x^2} + \frac{2}{x} \cr
& = \frac{1}{{12}}{\left( {{x^2} + \frac{2}{x}} \right)^6} + C \cr} $$
