Question

不定积分 x^2/x^2-3x+2 dx

Answer 1

原式=不定积分（x^2）/(x^2-3x+2)dx
=不定积分（x^2-3x+2+3x-2）/(x^2-3x+2)dx
=不定积分1dx+不定积分[3(x-1)+1]/(x-2)(x-1)dx
=x+不定积分3/(x-2)dx+不定积分1/(x-2)(x-1)dx
=x+3ln(x-2)+ln[(x-2)/(x-1)]+C
=x+ln[(x-2)^4/(x-1)]+C

Answer 2

∫ x^2/(x^2-3x+2) dx
=∫ [1 + (3x-2)/(x^2-3x+2) ]dx
=x + (3/2)∫dln(x^2-3x+2) + (5/2)∫ 1/(x^2-3x+2) dx
=x + (3/2)ln|x^2-3x+2| + (5/2)∫ 1/(x^2-3x+2) dx
=x + (3/2)ln|x^2-3x+2| + (5/2)∫ [1/(x-2) - 1/(x-1)] dx
=x + (3/2)ln|x^2-3x+2| + (5/2)ln|(x-2)/(x-1)| + C

Answer 3

你好：
x^2/（x^2-3x+2)=2x/x-2 - x/x-1

分成两部分分别作积分：

2x/x-2 根据牛顿莱布尼兹公式 得：2x ln（x-2）-2（x-2）ln（x-2）-ln（x-2）
=2（x-1）ln（x-2) - 2（x-2)

同理：x/x-1积分