integral of (5u-1)(3u^3+2)

解

\int (5 u - 1) (3 u^{3} + 2) d u

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

解答ステップ

\int (5 u - 1) (3 u^{3} + 2) d u

拡張

(5 u - 1) (3 u^{3} + 2) : 15 u^{4} + 10 u - 3 u^{3} - 2

= \int 15 u^{4} + 10 u - 3 u^{3} - 2 d u

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 15 u^{4} d u + \int 10 u d u - \int 3 u^{3} d u - \int 2 d u

\int 15 u^{4} d u = 3 u^{5}

\int 10 u d u = 5 u^{2}

\int 3 u^{3} d u = \frac{3 u ^{4}}{4}

\int 2 d u = 2 u

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

定数を解答に追加する

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