integral of (7x-1)(2x^2+1)

解

\int (7 x - 1) (2 x^{2} + 1) d x

\frac{7 x ^{4}}{2} + \frac{7 x ^{2}}{2} - \frac{2 x ^{3}}{3} - x + C

解答ステップ

\int (7 x - 1) (2 x^{2} + 1) d x

拡張

(7 x - 1) (2 x^{2} + 1) : 14 x^{3} + 7 x - 2 x^{2} - 1

= \int 14 x^{3} + 7 x - 2 x^{2} - 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 14 x^{3} d x + \int 7 x d x - \int 2 x^{2} d x - \int 1 d x

\int 14 x^{3} d x = \frac{7 x ^{4}}{2}

\int 7 x d x = \frac{7 x ^{2}}{2}

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

\int 1 d x = x

= \frac{7 x ^{4}}{2} + \frac{7 x ^{2}}{2} - \frac{2 x ^{3}}{3} - x

定数を解答に追加する

= \frac{7 x ^{4}}{2} + \frac{7 x ^{2}}{2} - \frac{2 x ^{3}}{3} - x + C