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

解

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

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

解答ステップ

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

拡張

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

= \int 2 x^{2} + x - 1 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 x d x - \int 1 d x

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

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

\int 1 d x = x

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

定数を解答に追加する

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