integral of (2x+1)(3-x)

解

\int (2 x + 1) (3 - x) d x

- \frac{2 x ^{3}}{3} + \frac{5 x ^{2}}{2} + 3 x + C

解答ステップ

\int (2 x + 1) (3 - x) d x

拡張

(2 x + 1) (3 - x) : - 2 x^{2} + 5 x + 3

= \int - 2 x^{2} + 5 x + 3 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 2 x^{2} d x + \int 5 x d x + \int 3 d x

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 3 d x = 3 x

= - \frac{2 x ^{3}}{3} + \frac{5 x ^{2}}{2} + 3 x

定数を解答に追加する

= - \frac{2 x ^{3}}{3} + \frac{5 x ^{2}}{2} + 3 x + C