求不定积分 求∫du/(u^3-1)
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解;
1/(u^3-1)
=1/[(u-1)(u^2+u+1)]
=A/(u-1)+(Bu+C)/(u^2+u+1)
用待定系数法求得:
A=1/3,B=-1/3,C=-2/3
∫du/(u^3-1)
=积分:[1/(3(u-1)-1/3(u+2)/(u^2+u+1)]du
=积分:[1/3(u-1)d(u-1)]-1/3[积分:1/2d(u^2+u+1)/(u^2+u+1)+3/2/[(u+1/2)^2(根号3/2)^2]d(u+1/2)]
=1/3ln|u-1|-1/6ln|u^2+u+1|-1/2*2/根号3arctan(u+1/2)/(根号3/2)+C
=1/3ln|u-1|-1/6ln|u^+u+1|-1/根号3arctan[(2u+1)/根号3]+C
(C为常数)
1/(u^3-1)
=1/[(u-1)(u^2+u+1)]
=A/(u-1)+(Bu+C)/(u^2+u+1)
用待定系数法求得:
A=1/3,B=-1/3,C=-2/3
∫du/(u^3-1)
=积分:[1/(3(u-1)-1/3(u+2)/(u^2+u+1)]du
=积分:[1/3(u-1)d(u-1)]-1/3[积分:1/2d(u^2+u+1)/(u^2+u+1)+3/2/[(u+1/2)^2(根号3/2)^2]d(u+1/2)]
=1/3ln|u-1|-1/6ln|u^2+u+1|-1/2*2/根号3arctan(u+1/2)/(根号3/2)+C
=1/3ln|u-1|-1/6ln|u^+u+1|-1/根号3arctan[(2u+1)/根号3]+C
(C为常数)
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