求有理函数不定积分 x/x^2-2x+2 dx
求有理函数不定积分 x/x^2-2x+2 dx
数学人气:412 ℃时间:2020-03-30 06:45:03
优质解答
∫x/(x^2-2x+2) dx=∫(x-1+1)/(x^2-2x+2) dx=∫(x-1)/(x^2-2x+2) dx+∫1/(x^2-2x+2) dx=1/2∫1/(x^2-2x+2) d(x^2-2x+2)+∫1/[(x-1)^2+1] dx=1/2ln(x^2-2x+2)+arctan(x-1)+C
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