integral of (5x+1)(2x-2)

解

\int (5 x + 1) (2 x - 2) d x

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

解答ステップ

\int (5 x + 1) (2 x - 2) d x

(5 x + 1) (2 x - 2) = 2 (5 x + 1) (x - 1)

= \int 2 (5 x + 1) (x - 1) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int (5 x + 1) (x - 1) d x

拡張

(5 x + 1) (x - 1) : 5 x^{2} - 4 x - 1

= 2 \cdot \int 5 x^{2} - 4 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

= 2 (\int 5 x^{2} d x - \int 4 x d x - \int 1 d x)

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

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

\int 1 d x = x

= 2 (\frac{5 x ^{3}}{3} - 2 x^{2} - x)

定数を解答に追加する

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