integral of (3m+4)^2

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

\int 16 d m = 16 m

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

定数を解答に追加する

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