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