∫1/(x^3-1)dx
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∫1/(x^3-1)dx
满意加二十分
满意加二十分
![∫1/(x^3-1)dx](/uploads/image/z/9066209-41-9.jpg?t=%E2%88%AB1%2F%28x%5E3-1%29dx)
x³-1=(x-1)(x²+x+1)
1/(x³-1)=A/(x-1)+(Bx+C)/(x²+x+1)
=1/3(x-1)-(x+2)/3(x²+x+1)
=1/3(x-1)-(2x+1)/6(x²+x+1)-1/2(x²+x+1)
∫1/(x³-1)dx
=∫dx/(x-1)[(x+½)²+¾]
=∫dx/3(x-1) - ∫(2x+1)dx/6(x²+x+1)
- (1/√3)∫d[(2x+1)/√3]/{[(2x+1)/√3]²+1}
=(⅓)ln|x-1|+(1/6)ln|x²+x+1|-(1/√3)arctan[(2x+1)/√3]+C
1/(x³-1)=A/(x-1)+(Bx+C)/(x²+x+1)
=1/3(x-1)-(x+2)/3(x²+x+1)
=1/3(x-1)-(2x+1)/6(x²+x+1)-1/2(x²+x+1)
∫1/(x³-1)dx
=∫dx/(x-1)[(x+½)²+¾]
=∫dx/3(x-1) - ∫(2x+1)dx/6(x²+x+1)
- (1/√3)∫d[(2x+1)/√3]/{[(2x+1)/√3]²+1}
=(⅓)ln|x-1|+(1/6)ln|x²+x+1|-(1/√3)arctan[(2x+1)/√3]+C