integral of (2x^2-x-1)(3x+4)

解

\int (2 x^{2} - x - 1) (3 x + 4) d x

\frac{3 x ^{4}}{2} + \frac{5 x ^{3}}{3} - \frac{7 x ^{2}}{2} - 4 x + C

解答ステップ

\int (2 x^{2} - x - 1) (3 x + 4) d x

拡張

(2 x^{2} - x - 1) (3 x + 4) : 6 x^{3} + 5 x^{2} - 7 x - 4

= \int 6 x^{3} + 5 x^{2} - 7 x - 4 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 6 x^{3} d x + \int 5 x^{2} d x - \int 7 x d x - \int 4 d x

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

\int 4 d x = 4 x

= \frac{3 x ^{4}}{2} + \frac{5 x ^{3}}{3} - \frac{7 x ^{2}}{2} - 4 x

定数を解答に追加する

= \frac{3 x ^{4}}{2} + \frac{5 x ^{3}}{3} - \frac{7 x ^{2}}{2} - 4 x + C