integral of (3s+4)^2

解

\int (3 s + 4)^{2} d s

3 s^{3} + 12 s^{2} + 16 s + C

解答ステップ

\int (3 s + 4)^{2} d s

拡張

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

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

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 s^{2} d s + \int 24 s d s + \int 16 d s

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

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

\int 16 d s = 16 s

= 3 s^{3} + 12 s^{2} + 16 s

定数を解答に追加する

= 3 s^{3} + 12 s^{2} + 16 s + C