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

解答

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

\frac{x ^{5}}{5} - \frac{5 x ^{4}}{4} + x^{3} + \frac{9 x ^{2}}{2} + C

求解步骤

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

乘开

x (x - 3)^{2} (x + 1) : x^{4} - 5 x^{3} + 3 x^{2} + 9 x

= \int x^{4} - 5 x^{3} + 3 x^{2} + 9 x d x

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 x^{4} d x - \int 5 x^{3} d x + \int 3 x^{2} d x + \int 9 x d x

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

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

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

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

= \frac{x ^{5}}{5} - \frac{5 x ^{4}}{4} + x^{3} + \frac{9 x ^{2}}{2}

解答补常数

= \frac{x ^{5}}{5} - \frac{5 x ^{4}}{4} + x^{3} + \frac{9 x ^{2}}{2} + C