integral of (3+2x)^2

解

\int (3 + 2 x)^{2} d x

9 x + 6 x^{2} + \frac{4 x ^{3}}{3} + C

解答ステップ

\int (3 + 2 x)^{2} d x

拡張

(3 + 2 x)^{2} : 9 + 12 x + 4 x^{2}

= \int 9 + 12 x + 4 x^{2} 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 d x + \int 12 x d x + \int 4 x^{2} d x

\int 9 d x = 9 x

\int 12 x d x = 6 x^{2}

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

= 9 x + 6 x^{2} + \frac{4 x ^{3}}{3}

定数を解答に追加する

= 9 x + 6 x^{2} + \frac{4 x ^{3}}{3} + C