Answer
$F\left( x \right) = \root 4 \of 5 x + \frac{{4{x^{5/4}}}}{5} + C$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \root 4 \of 5 + \root 4 \of x \cr
& {\text{First use the radical property }}\root 4 \of m = {m^{1/4}} \cr
& {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr
& {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr
& {\text{Function: }}g\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}g\left( x \right) + h\left( x \right) \cr
& {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr
& F\left( x \right) = \root 4 \of 5 + {x^{1/4}} \cr
& F\left( x \right) = \root 4 \of 5 x + \frac{{{x^{1/4 + 1}}}}{{1/4 + 1}} + C \cr
& F\left( x \right) = \root 4 \of 5 x + \frac{{{x^{5/4}}}}{{5/4}} + C \cr
& F\left( x \right) = \root 4 \of 5 x + \frac{{4{x^{5/4}}}}{5} + C \cr} $$
