求dx/[(x-1)(x^2+4x+9)]不定积分?_数学解答

求dx/[(x-1)(x^2+4x+9)]不定积分?

人气：261 ℃　时间：2020-04-04 03:17:05

解答

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 - C1 = (A + B)x² + (4A - B + C)x ...