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不定积分 x^2/x^2-3x+2 dx
人气:171 ℃ 时间:2020-02-05 15:46:48
解答
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
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