Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 362: 23

$F\left( r \right) = 4{\tan ^{ - 1}}r - \frac{{5{r^{9/5}}}}{9} + C$

$$\eqalign{ & f\left( r \right) = \frac{4}{{1 + {r^2}}} - \root 5 \of {{r^4}} \cr & {\text{Rewrite the function using radical properties}} \cr & f\left( r \right) = \frac{4}{{1 + {r^2}}} - {r^{4/5}} \cr & f\left( r \right) = 4\left( {\frac{1}{{1 + {r^2}}}} \right) - {r^{4/5}} \cr & {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right){\text{ }} \cr & {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}\frac{1}{{1 + {x^2}}} \to {\text{Particular antiderivative: ta}}{{\text{n}}^{ - 1}}x \cr & {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{x^{n + 1}}}}{{n + 1}} \cr & {\text{Function: }}f\left( x \right) + h\left( x \right) \to {\text{Particular antiderivative: }}f\left( x \right) + h\left( x \right) \cr & {\text{And applying the Theorem 1}},{\text{ we obtain}} \cr & f\left( r \right) = 4\left( {\frac{1}{{1 + {r^2}}}} \right) - {r^{4/5}} \cr & F\left( r \right) = 4{\tan ^{ - 1}}r - \frac{{{r^{4/5 + 1}}}}{{4/5 + 1}} + C \cr & F\left( r \right) = 4{\tan ^{ - 1}}r - \frac{{{r^{9/5}}}}{{9/5}} + C \cr & F\left( r \right) = 4{\tan ^{ - 1}}r - \frac{{5{r^{9/5}}}}{9} + C \cr & \cr & {\text{Checking the answer by differentiation}} \cr & F'\left( r \right) = 4\frac{d}{{dr}}\left[ {{{\tan }^{ - 1}}r} \right] + \frac{5}{9}\frac{d}{{dr}}\left[ {{r^{9/5}}} \right] + \frac{d}{{dr}}\left[ C \right] \cr & F'\left( r \right) = 4\left( {\frac{1}{{1 + {r^2}}}} \right) + \frac{5}{9}\left( {\frac{9}{5}{r^{4/5}}} \right) + 0 \cr & F'\left( r \right) = \frac{4}{{1 + {r^2}}} + {r^{4/5}} \cr & F'\left( r \right) = \frac{4}{{1 + {r^2}}} + \root 5 \of {{r^4}} \cr} $$