求dx/[(x-1)(x^2+4x+9)]不定积分?

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/04/28 16:50:42

令1/[(x - 1)(x² + 4x + 9)]
= A/(x - 1) + (Bx + C)/(x² + 4x + 9)
==> 1 = A(x² + 4x + 9) + (Bx + C)(x - 1)
1 = Ax² + 4Ax + 9A + Bx² + Cx - Bx - C
1 = (A + B)x² + (4A - B + C)x + (9A - C)
{ A + B = 0 ==> B = - A
{ 4A - B + C = 0 ==> 4A + A + C = 0 ==> 5A + C = 0
{ 9A - C = 1
(5A + C) + (9A - C) = 0 + 1
14A = 1 ==> A = 1/14
B = - 1/14
C = - 5(1/14) = - 5/14
∫ 1/[(x - 1)(x² + 4x + 9)] dx
= (1/14)∫ dx/(x - 1) - (1/14)∫ (x + 5)/(x² + 4x + 9) dx,d(x² + 4x + 9) = (2x + 4) dx
= (1/14)ln|x - 1| - (1/14)∫ {(1/2)[(2x + 4) - 4] + 5}/(x² + 4x + 9) dx
= (1/14)ln|x - 1| - (1/28)∫ (2x + 4)/(x² + 4x + 9) dx - (3/14)∫ dx/[(x + 2)² + 5]
= (1/14)ln|x - 1| - (1/28)ln|x² + 4x + 9| - [3/(14√5)]arctan[(x + 2)/√5] + C

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