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

解

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

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

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

拡張

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

= 2 \cdot \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

= 2 (\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}

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

定数を解答に追加する

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