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求不定积分∫[1/(1+x^3)]dx
人气:195 ℃ 时间:2020-03-23 10:04:45
解答
1+x^3=(x+1)(x^2-x+1)用待定系数法:A/(x+1)+(Bx+c)/(x^2-x+1)=1/(x+1)(x^2-x+1)得A=1/3,B=-1/3,C=2/3所以∫[1/(1+x^3)]dx =1/3∫(1/(x+1))dx-1/3∫((x-2)/(x^2-x+1))dx其中1/3∫(1/(x+1))dx=1/3ln|x+1|+c因为d(x^2...
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