一道高数题不定积分的谢谢回答 ∫3/（x^3+1)dx
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∫3/(x^3+1)dx
=∫(x^2-x+1)+(x+1)-(x^2-1)/(x+1)(x^2-x+1)
=∫带租dx/(x+1)+∫dx/(x^2-x+1)-∫(x-1)/(x^2-x+1)dx
=∫dx/(x+1)-1/野铅2∫(2x-1-1)/颂行好(x^2-x+1)+∫dx/(x^2-x+1)
=∫dx/(x+1)-1/2∫(x^2-x+1)/(x^2-x+1)+3/2∫dx/(x^2-x+1)
=ln|x+1|-(1/2)ln|x^2-x+1|+√3 arctan[(2x-1)/√3]+c
=∫(x^2-x+1)+(x+1)-(x^2-1)/(x+1)(x^2-x+1)
=∫带租dx/(x+1)+∫dx/(x^2-x+1)-∫(x-1)/(x^2-x+1)dx
=∫dx/(x+1)-1/野铅2∫(2x-1-1)/颂行好(x^2-x+1)+∫dx/(x^2-x+1)
=∫dx/(x+1)-1/2∫(x^2-x+1)/(x^2-x+1)+3/2∫dx/(x^2-x+1)
=ln|x+1|-(1/2)ln|x^2-x+1|+√3 arctan[(2x-1)/√3]+c
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