integral of (-x^2-3x+6)^2

解答

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

\frac{x ^{5}}{5} + \frac{3 x ^{4}}{2} - x^{3} - 18 x^{2} + 36 x + C

求解步骤

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

因式分解

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

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

化简

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

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

乘开

(x^{2} + 3 x - 6)^{2} : x^{4} + 6 x^{3} - 3 x^{2} - 36 x + 36

= \int x^{4} + 6 x^{3} - 3 x^{2} - 36 x + 36 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 6 x^{3} d x - \int 3 x^{2} d x - \int 36 x d x + \int 36 d x

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

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

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

\int 36 x d x = 18 x^{2}

\int 36 d x = 36 x

= \frac{x ^{5}}{5} + \frac{3 x ^{4}}{2} - x^{3} - 18 x^{2} + 36 x

解答补常数

= \frac{x ^{5}}{5} + \frac{3 x ^{4}}{2} - x^{3} - 18 x^{2} + 36 x + C

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 0

d er i v a t i v e \frac{2 ^{x}}{3 x - 1}

n = 1 \sum \infty \frac{5 n ln ( n )}{6 ^{n}}

d er i v a t i v e \frac{5 + csc ( x )}{9 - csc ( x )}

(1 + t^{2}) \frac{d y}{d t} + 2 t y = t