integral of x(x^2-4)^2

解

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

\frac{x ^{6}}{6} - 2 x^{4} + 8 x^{2} + C

解答ステップ

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

拡張

x (x^{2} - 4)^{2} : x^{5} - 8 x^{3} + 16 x

= \int x^{5} - 8 x^{3} + 16 x 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^{5} d x - \int 8 x^{3} d x + \int 16 x d x

\int x^{5} d x = \frac{x ^{6}}{6}

\int 8 x^{3} d x = 2 x^{4}

\int 16 x d x = 8 x^{2}

= \frac{x ^{6}}{6} - 2 x^{4} + 8 x^{2}

定数を解答に追加する

= \frac{x ^{6}}{6} - 2 x^{4} + 8 x^{2} + C