Answer
$$2{x^{1/2}} - \frac{{15{x^{12/5}}}}{{12}} + \frac{1}{9}x + C$$
Work Step by Step
$$\eqalign{
& \int {\left[ {{x^{ - 1/2}} - 3{x^{7/5}} + \frac{1}{9}} \right]dx} \cr
& {\text{By the theorem 4}}{\text{.2}}{\text{.3 }}\left( b \right){\text{ and }}\left( c \right) \cr
& = \int {{x^{ - 1/2}}dx} - \int {3{x^{7/5}}} dx + \int {\frac{1}{9}} dx \cr
& {\text{By the theorem 4}}{\text{.2}}{\text{.3 }}\left( a \right) \cr
& = \int {{x^{ - 1/2}}dx} - 3\int {{x^{7/5}}} dx + \frac{1}{9}\int {dx} \cr
& {\text{applying the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{{{x^{1/2}}}}{{1/2}} - 3\left( {\frac{{{x^{12/5}}}}{{12/5}}} \right) + \frac{1}{9}x + C \cr
& {\text{Simplifying}} \cr
& = 2{x^{1/2}} - \frac{{15{x^{12/5}}}}{{12}} + \frac{1}{9}x + C \cr} $$
