求∫1/(x^4-1)的不定积分
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∫1/(x^4-1)dx=∫1/[(x?-1)(x?+1)]dx=(1/2)∫[1/(x?-1)]-[1/(x?+1)]dx=(1/2)∫[1/(x?-1)]dx-(1/2)∫[1/(x?+1)]dx=(1/4)∫[1/(x-1)]-[1/(x+1)]dx-(1/2)arctanx=(1/4)∫[1/(x-1)]dx-(1/4)∫[1/(x+1)]dx-(1/2)arctanx=(1/4)ln|x-1|-(1/4)ln|x+1|-(1/2)arctanx=(1/4)ln(|x-1|/|x+1|)-(1/2)arctanx
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