Answer
$$\frac{{{x^2}}}{2} - \frac{2}{x} + \frac{1}{{3{x^3}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^5} + 2{x^2} - 1}}{{{x^4}}}} dx \cr
& {\text{distribute numerator}} \cr
& = \int {\left( {\frac{{{x^5}}}{{{x^4}}} + \frac{{2{x^2}}}{{{x^4}}} - \frac{1}{{{x^4}}}} \right)} dx \cr
& \frac{{{x^m}}}{{{x^n}}} = {x^{m - n}} \cr
& = \int {\left( {{x^{5 - 4}} + 2{x^{2 - 4}} - {x^{ - 4}}} \right)} dx \cr
& = \int {\left( {x + 2{x^{ - 2}} - {x^{ - 4}}} \right)} dx \cr
& {\text{power rule}} \cr
& = \frac{{{x^2}}}{2} + \frac{{2{x^{ - 1}}}}{{ - 1}} - \frac{{{x^{ - 3}}}}{{ - 3}} + C \cr
& = \frac{{{x^2}}}{2} - \frac{2}{x} + \frac{1}{{3{x^3}}} + C \cr
& \cr
& {\text{check by differentiation}} \cr
& = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{2} + \frac{{2{x^{ - 1}}}}{{ - 1}} - \frac{{{x^{ - 3}}}}{{ - 3}} + C} \right] \cr
& = \frac{{2x}}{2} - \frac{{2{x^{ - 2}}}}{{ - 1}} - \frac{{ - 3{x^{ - 4}}}}{{ - 3}} + 0 \cr
& = x + \frac{2}{{{x^2}}} - \frac{1}{{{x^4}}} \cr} $$
