1/(x^2+1)(x^2+x+1)的不定积分

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∫1/(x²+1)(x²+x+1)=∫[(-x)/(x²+1)+(x+1)/(x²+x+1)]=∫[(-x)/(x²+1)+(x+1)/(x²+x+1)]=(-1/2)ln(x²+1)+∫(x+1)/(x²+x+1)∫(x+1)/(x²+x+1)令x+1/2＝t,则：dx＝dt.∴∫[...

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