Answer
$$\frac{{{x^3}}}{3} + \frac{4}{{{x^{1/2}}}} - \frac{2}{x} + C$$
Work Step by Step
$$\eqalign{
& {\text{Find }}\int {\frac{{{x^4} - 2\sqrt x + 2}}{{{x^2}}}dx} \cr
& {\text{distribute}} \cr
& = \int {\left( {\frac{{{x^4}}}{{{x^2}}} - \frac{{2\sqrt x }}{{{x^2}}} + \frac{2}{{{x^2}}}} \right)dx} \cr
& = \int {\left( {{x^2} - 2{x^{ - 3/2}} + 2{x^{ - 2}}} \right)dx} \cr
& {\text{use }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{{{x^{2 + 1}}}}{{2 + 1}} - 2\left( {\frac{{{x^{ - 3/2 + 1}}}}{{ - 3/2 + 1}}} \right) + 2\left( {\frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right) + C \cr
& {\text{simplify}} \cr
& = \frac{{{x^3}}}{3} - 2\left( {\frac{{{x^{ - 1/2}}}}{{ - 1/2}}} \right) + 2\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right) + C \cr
& = \frac{{{x^3}}}{3} + \frac{4}{{{x^{1/2}}}} - \frac{2}{x} + C \cr} $$
