integral of [(3-x^2)^3]

解

\int [(3 - x^{2})^{3}] d x

27 x - 9 x^{3} + \frac{9 x ^{5}}{5} - \frac{x ^{7}}{7} + C

解答ステップ

\int (3 - x^{2})^{3} d x

拡張

(3 - x^{2})^{3} : 27 - 27 x^{2} + 9 x^{4} - x^{6}

= \int 27 - 27 x^{2} + 9 x^{4} - x^{6} d x

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 27 d x - \int 27 x^{2} d x + \int 9 x^{4} d x - \int x^{6} d x

\int 27 d x = 27 x

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

\int 9 x^{4} d x = \frac{9 x ^{5}}{5}

\int x^{6} d x = \frac{x ^{7}}{7}

= 27 x - 9 x^{3} + \frac{9 x ^{5}}{5} - \frac{x ^{7}}{7}

定数を解答に追加する

= 27 x - 9 x^{3} + \frac{9 x ^{5}}{5} - \frac{x ^{7}}{7} + C