integral of (x^3-3)^2

解答

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

\frac{x ^{7}}{7} - \frac{3 x ^{4}}{2} + 9 x + C

求解步骤

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

乘开

(x^{3} - 3)^{2} : x^{6} - 6 x^{3} + 9

= \int x^{6} - 6 x^{3} + 9 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^{6} d x - \int 6 x^{3} d x + \int 9 d x

\int x^{6} d x = \frac{x ^{7}}{7}

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

\int 9 d x = 9 x

= \frac{x ^{7}}{7} - \frac{3 x ^{4}}{2} + 9 x

解答补常数

= \frac{x ^{7}}{7} - \frac{3 x ^{4}}{2} + 9 x + C

\int \frac{cos ( x )}{sin ^{8} ( x )} d x

x^{2} d x + 2 y d y = 0, y (0) = 2

d er i v a t i v e (\frac{x}{3} + 9)^{\frac{3}{2}}

\int_{0}^{- 1} \frac{x ^{3} + 8}{x + 2} d x

\frac{d y}{d x} + \frac{y}{x} = x^{3} y^{2}