integral of (x^2-4)^3

解

\int (x^{2} - 4)^{3} d x

\frac{x ^{7}}{7} - \frac{12 x ^{5}}{5} + 16 x^{3} - 64 x + C

解答ステップ

\int (x^{2} - 4)^{3} d x

拡張

(x^{2} - 4)^{3} : x^{6} - 12 x^{4} + 48 x^{2} - 64

= \int x^{6} - 12 x^{4} + 48 x^{2} - 64 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 x^{6} d x - \int 12 x^{4} d x + \int 48 x^{2} d x - \int 64 d x

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

\int 12 x^{4} d x = \frac{12 x ^{5}}{5}

\int 48 x^{2} d x = 16 x^{3}

\int 64 d x = 64 x

= \frac{x ^{7}}{7} - \frac{12 x ^{5}}{5} + 16 x^{3} - 64 x

定数を解答に追加する

= \frac{x ^{7}}{7} - \frac{12 x ^{5}}{5} + 16 x^{3} - 64 x + C