Answer
$F\left( x \right) = 5x - 6\ln \left| x \right| - \frac{4}{x} + C$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{5{x^2} - 6x + 4}}{{{x^2}}},{\text{ }}x > 0 \cr
& {\text{Distribute the numerator}} \cr
& f\left( x \right) = \frac{{5{x^2}}}{{{x^2}}} - \frac{{6x}}{{{x^2}}} + \frac{4}{{{x^2}}} \cr
& f\left( x \right) = 5 - \frac{6}{x} + \frac{4}{{{x^2}}} \cr
& {\text{Rewrite the function using property of exponents }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr
& f\left( x \right) = 5 - \frac{6}{x} + 4{x^{ - 2}} \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{x^{n + 1}}}}{{n + 1}} \cr
& {\text{Function: }}\frac{1}{x} \to {\text{Particular antiderivative: }}\ln \left| x \right| \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& f\left( x \right) = 5 - \frac{6}{x} + 4{x^{ - 2}} \cr
& F\left( x \right) = 5x - 6\ln \left| x \right| + 4\left( {\frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right) + C \cr
& F\left( x \right) = 5x - 6\ln \left| x \right| + 4\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right) + C \cr
& F\left( x \right) = 5x - 6\ln \left| x \right| - \frac{4}{x} + C \cr
& \cr
& {\text{Checking the answer by differentiation}} \cr
& F'\left( x \right) = \frac{d}{{dx}}\left[ {5x} \right] - \frac{d}{{dx}}\left[ {6\ln \left| x \right|} \right] - \frac{d}{{dx}}\left[ {\frac{4}{x}} \right] + \frac{d}{{dx}}\left[ C \right] \cr
& F'\left( x \right) = 5 - 6\left( {\frac{1}{x}} \right) - \left( { - \frac{4}{{{x^2}}}} \right) + 0 \cr
& F'\left( x \right) = 5 - \frac{6}{x} + \frac{4}{{{x^2}}} \cr} $$
