Answer
$F\left( x \right) = \frac{2}{5}\ln \left| x \right| + \frac{3}{x} + C$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{2}{{5x}} - \frac{3}{{{x^2}}} \cr
& {\text{Rewrite the function using property of exponents }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr
& f\left( x \right) = \frac{2}{{5x}} - 3{x^{ - 2}} \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{x^{n + 1}}}}{{n + 1}} \cr
& {\text{Function: }}\frac{1}{x} \to {\text{Particular antiderivative: }}\ln \left| x \right| \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& F\left( x \right) = \frac{2}{5}\ln \left| x \right| - 3\left( {\frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}}} \right) + C \cr
& F\left( x \right) = \frac{2}{5}\ln \left| x \right| - 3\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right) + C \cr
& F\left( x \right) = \frac{2}{5}\ln \left| x \right| + \frac{3}{x} + C \cr
& \cr
& {\text{Checking the answer by differentiation}} \cr
& F'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{2}{5}\ln \left| x \right|} \right] + \frac{d}{{dx}}\left[ {\frac{3}{x}} \right] + \frac{d}{{dx}}\left[ C \right] \cr
& F'\left( x \right) = \frac{2}{5}\left( {\frac{1}{x}} \right) + 3\left( { - \frac{1}{{{x^2}}}} \right) + 0 \cr
& F'\left( x \right) = \frac{2}{{5x}} - \frac{3}{{{x^2}}} \cr} $$
