integral of (3x-4)^2

解

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

3 x^{3} - 12 x^{2} + 16 x + C

解答ステップ

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

拡張

(3 x - 4)^{2} : 9 x^{2} - 24 x + 16

= \int 9 x^{2} - 24 x + 16 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 9 x^{2} d x - \int 24 x d x + \int 16 d x

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

\int 24 x d x = 12 x^{2}

\int 16 d x = 16 x

= 3 x^{3} - 12 x^{2} + 16 x

定数を解答に追加する

= 3 x^{3} - 12 x^{2} + 16 x + C