integral of (u+4)(2u+1)

解答

\int (u + 4) (2 u + 1) d u

\frac{2 u ^{3}}{3} + \frac{9 u ^{2}}{2} + 4 u + C

求解步骤

\int (u + 4) (2 u + 1) d u

乘开

(u + 4) (2 u + 1) : 2 u^{2} + 9 u + 4

= \int 2 u^{2} + 9 u + 4 d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 u^{2} d u + \int 9 u d u + \int 4 d u

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

\int 9 u d u = \frac{9 u ^{2}}{2}

\int 4 d u = 4 u

= \frac{2 u ^{3}}{3} + \frac{9 u ^{2}}{2} + 4 u

解答补常数

= \frac{2 u ^{3}}{3} + \frac{9 u ^{2}}{2} + 4 u + C