求∫dx/x^3+x
要考试了,帮帮啦各位,谢谢!!!!!
因为:1/(x^3+x)=1/[x*(x^2+1)] 设1/(x^3+x)=(A/x)+[Bx/(x^2+1)] =[A(x^2+1)+Bx^2]/[x*(x^2+1)] =[(A+B)x^2+A]/[x*(x^2+1)] 所以: A+B=0 A=1 解得:A=1,B=-1 所以:原不定积分=∫[(1/x)-x/(x^2+1)]dx =∫(1/x)dx-∫[x/(x^2+1)]dx =ln|x|-(1/2)∫[1/(x^2+1)]d(x^2+1) =ln|x|-(1/2)ln(x^2+1)+C.
原式=ln|x|-(1/2)ln(x^2+1)+C
∫ dx/(x³+x) =∫[1/(x³+x)]dx =∫{1/[x² (x+1)]}dx =∫{(-x+1)/(x²) + 1/(x+1)}dx =∫ (-x+1)/(x²) dx+∫ dx/(x+1) ==-(1/2)∫(2x)/ x²+∫dx/(x²)+∫ dx/(x+1) =-(1/2)∫d(x²)/ (x²)dx+∫dx/(x²)+ ln|x+1| =-(1/2)ln|x²|-(1/x)+ln|x+1|+C
用待定系数法做将它分成两个分子为一的分数之和,就可以用函数表示啦,微积分上有啊
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