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

解

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

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

解答ステップ

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

拡張

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

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

\int x^{4} d x = \frac{x ^{5}}{5}

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

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

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

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

定数を解答に追加する

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