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求有理函数的不定积分§X/(X^3+1)dx
人气:388 ℃ 时间:2020-04-09 02:22:01
解答
∫x/(x³+1) dx
=1/3*∫(x+1)/(x²-x+1)-1/3*∫1/(x+1)
=1/6*∫(2x-1)/(x²-x+1)+1/2*∫1/(x²-x+1)
=1/6*ln(x²-x+1)+1/2*∫1/[(x-1/2)²+3/4]
=1/6*ln(x²-x+1)-1/3*ln(x+1)+1/√3*arctan[(2x-1)/√3]
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