如何用因式分解法解题？

Question

如题所述

Answer 1

解答过程如下：

先分解因式：

∫ 1/(x³ + 1) dx = ∫ 1/[(x + 1)(x² - x + 1)] dx

= ∫ A/(x + 1) dx + ∫ (Bx + C)/(x² - x + 1) dx

1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + C

1 = (A + B)x² + (- A + B + C)x + (A + C)

{ A + B = 0

{ - A + B + C = 0

{ A + C = 1

(A + B) - (- A + B + C) = 0 ==> { 2A - C = 0

{A + C = 1

(2A - C) + (A + C) = 1 ==> 3A = 1 ==> A = 1/3

B = - 1/3,C = 2/3

原式 = (1/3)∫ dx/(x + 1) - (1/3)∫ (x - 2)/(x² - x + 1) dx

= (1/3)∫ d(x + 1)/(x + 1) - (1/3)∫ [(2x - 1)/2 - 3/2]/(x² - x + 1) dx

= (1/3)ln(x + 1) - (1/6)∫ d(x² - x + 1)/(x² - x +1) + (1/2)∫ d(x - 1/2)/[(x - 1/2)² + 3/4]

= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/2)(2/√3)arctan[(x - 1/2) * 2/√3] + C

= (1/3)ln(x + 1) - (1/6)ln(x² - x + 1) + (1/√3)arctan(2x/√3 - 1/√3) + C

扩展资料

分部积分：

(uv)'=u'v+uv'

得：u'v=(uv)'-uv'

两边积分得：∫ u'v dx=∫ (uv)' dx - ∫ uv' dx

即：∫ u'v dx = uv - ∫ uv' d,这就是分部积分公式

也可简写为：∫ v du = uv - ∫ u dv

常用积分公式：

1）∫0dx=c 

2）∫x^udx=(x^(u+1))/(u+1)+c

3）∫1/xdx=ln|x|+c

4）∫a^xdx=(a^x)/lna+c

5）∫e^xdx=e^x+c

6）∫sinxdx=-cosx+c

7）∫cosxdx=sinx+c

8）∫1/(cosx)^2dx=tanx+c

9）∫1/(sinx)^2dx=-cotx+c

10）∫1/√（1-x^2) dx=arcsinx+c