Answer
$F\left( x \right) = - \frac{2}{{{x^5}}} - 2{e^x} + 3x + C$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{10}}{{{x^6}}} - 2{e^x} + 3 \cr
& {\text{Rewrite the function using property of exponents }}\frac{1}{{{x^n}}} = {x^{ - n}} \cr
& f\left( x \right) = 10{x^{ - 6}} - 2{e^x} + 3 \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{x^{n + 1}}}}{{n + 1}} \cr
& {\text{Function: }}{e^x} \to {\text{Particular antiderivative: }}{e^x} \cr
& {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& f\left( x \right) = 10{x^{ - 6}} - 2{e^x} + 3 \cr
& F\left( x \right) = 10\left( {\frac{{{x^{ - 6 + 1}}}}{{ - 6 + 1}}} \right) - 2\left( {{e^x}} \right) + 3\left( x \right) + C \cr
& F\left( x \right) = 10\left( {\frac{{{x^{ - 5}}}}{{ - 5}}} \right) - 2{e^x} + 3x + C \cr
& F\left( x \right) = - 2{x^{ - 5}} - 2{e^x} + 3x + C \cr
& F\left( x \right) = - \frac{2}{{{x^5}}} - 2{e^x} + 3x + C \cr
& \cr
& {\text{Checking the answer by differentiation}} \cr
& F'\left( x \right) = - \frac{d}{{dx}}\left[ {2{x^{ - 5}}} \right] - \frac{d}{{dx}}\left[ {2{e^x}} \right] + \frac{d}{{dx}}\left[ {3x} \right] + \frac{d}{{dx}}\left[ C \right] \cr
& F'\left( x \right) = - \left( { - 10{x^{ - 6}}} \right) - 2{e^x} + 3\left( 1 \right) + 0 \cr
& F'\left( x \right) = 10{x^{ - 6}} - 2{e^x} + 3 \cr
& F'\left( x \right) = \frac{{10}}{{{x^6}}} - 2{e^x} + 3 \cr} $$
