integral of (x+2)^2(1-x)^6

解答

\int (x + 2)^{2} (1 - x)^{6} d x

\frac{( x + 2 ) ^{9}}{9} - \frac{9 ( x + 2 ) ^{8}}{4} + \frac{135 ( x + 2 ) ^{7}}{7} - 90 (x + 2)^{6} + 243 (x + 2)^{5} - \frac{729 ( x + 2 ) ^{4}}{2} + 243 (x + 2)^{3} + C

求解步骤

\int (x + 2)^{2} (1 - x)^{6} d x

使用换元积分法

= \int u^{2} (- u + 3)^{6} d u

因式分解

- u + 3 : - (u - 3)

= \int u^{2} (- (u - 3))^{6} d u

(- (u - 3))^{6} = (u - 3)^{6}

= \int u^{2} (u - 3)^{6} d u

乘开

u^{2} (u - 3)^{6} : u^{8} - 18 u^{7} + 135 u^{6} - 540 u^{5} + 1215 u^{4} - 1458 u^{3} + 729 u^{2}

= \int u^{8} - 18 u^{7} + 135 u^{6} - 540 u^{5} + 1215 u^{4} - 1458 u^{3} + 729 u^{2} 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 u^{8} d u - \int 18 u^{7} d u + \int 135 u^{6} d u - \int 540 u^{5} d u + \int 1215 u^{4} d u - \int 1458 u^{3} d u + \int 729 u^{2} d u

\int u^{8} d u = \frac{u ^{9}}{9}

\int 18 u^{7} d u = \frac{9 u ^{8}}{4}

\int 135 u^{6} d u = \frac{135 u ^{7}}{7}

\int 540 u^{5} d u = 90 u^{6}

\int 1215 u^{4} d u = 243 u^{5}

\int 1458 u^{3} d u = \frac{729 u ^{4}}{2}

\int 729 u^{2} d u = 243 u^{3}

= \frac{u ^{9}}{9} - \frac{9 u ^{8}}{4} + \frac{135 u ^{7}}{7} - 90 u^{6} + 243 u^{5} - \frac{729 u ^{4}}{2} + 243 u^{3}

u = x + 2

代回

= \frac{( x + 2 ) ^{9}}{9} - \frac{9 ( x + 2 ) ^{8}}{4} + \frac{135 ( x + 2 ) ^{7}}{7} - 90 (x + 2)^{6} + 243 (x + 2)^{5} - \frac{729 ( x + 2 ) ^{4}}{2} + 243 (x + 2)^{3}

解答补常数

= \frac{( x + 2 ) ^{9}}{9} - \frac{9 ( x + 2 ) ^{8}}{4} + \frac{135 ( x + 2 ) ^{7}}{7} - 90 (x + 2)^{6} + 243 (x + 2)^{5} - \frac{729 ( x + 2 ) ^{4}}{2} + 243 (x + 2)^{3} + C

a re a 5 - x^{2}, sin (x), 0, 2

\frac{d ^{4}}{d x ^{4}} (8 e^{- 2 x})

\frac{d y}{d x} = (x - 2) e^{2 y}

t an g e n t y = (6 x^{2} - 6 x + 2) (1 + 2 x), at x = 1

\frac{\partial}{\partial y} (cos (x^{2} + y^{3}))