integral of (s^3+3)^2(1-s^2)^3

解

\int (s^{3} + 3)^{2} (1 - s^{2})^{3} d s

- \frac{8 s ^{7}}{7} - \frac{s ^{9}}{3} + \frac{3 s ^{11}}{11} - \frac{s ^{13}}{13} - \frac{3 s ^{10}}{5} + \frac{9 s ^{8}}{4} + \frac{27 s ^{5}}{5} + \frac{3 s ^{4}}{2} - 3 s^{6} - 9 s^{3} + 9 s + C

解答ステップ

\int (s^{3} + 3)^{2} (1 - s^{2})^{3} d s

拡張

(s^{3} + 3)^{2} (1 - s^{2})^{3} : - 8 s^{6} - 3 s^{8} + 3 s^{10} - s^{12} - 6 s^{9} + 18 s^{7} + 27 s^{4} + 6 s^{3} - 18 s^{5} - 27 s^{2} + 9

= \int - 8 s^{6} - 3 s^{8} + 3 s^{10} - s^{12} - 6 s^{9} + 18 s^{7} + 27 s^{4} + 6 s^{3} - 18 s^{5} - 27 s^{2} + 9 d s

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 8 s^{6} d s - \int 3 s^{8} d s + \int 3 s^{10} d s - \int s^{12} d s - \int 6 s^{9} d s + \int 18 s^{7} d s + \int 27 s^{4} d s + \int 6 s^{3} d s - \int 18 s^{5} d s - \int 27 s^{2} d s + \int 9 d s

\int 8 s^{6} d s = \frac{8 s ^{7}}{7}

\int 3 s^{8} d s = \frac{s ^{9}}{3}

\int 3 s^{10} d s = \frac{3 s ^{11}}{11}

\int s^{12} d s = \frac{s ^{13}}{13}

\int 6 s^{9} d s = \frac{3 s ^{10}}{5}

\int 18 s^{7} d s = \frac{9 s ^{8}}{4}

\int 27 s^{4} d s = \frac{27 s ^{5}}{5}

\int 6 s^{3} d s = \frac{3 s ^{4}}{2}

\int 18 s^{5} d s = 3 s^{6}

\int 27 s^{2} d s = 9 s^{3}

\int 9 d s = 9 s

= - \frac{8 s ^{7}}{7} - \frac{s ^{9}}{3} + \frac{3 s ^{11}}{11} - \frac{s ^{13}}{13} - \frac{3 s ^{10}}{5} + \frac{9 s ^{8}}{4} + \frac{27 s ^{5}}{5} + \frac{3 s ^{4}}{2} - 3 s^{6} - 9 s^{3} + 9 s

定数を解答に追加する

= - \frac{8 s ^{7}}{7} - \frac{s ^{9}}{3} + \frac{3 s ^{11}}{11} - \frac{s ^{13}}{13} - \frac{3 s ^{10}}{5} + \frac{9 s ^{8}}{4} + \frac{27 s ^{5}}{5} + \frac{3 s ^{4}}{2} - 3 s^{6} - 9 s^{3} + 9 s + C