Question

∫x/(x^2-x+1)dx用凑微分法怎么求？

Answer 1

x/(x^2-x+1) = (x -1/2) /(x^2-x+1) + (1/2) /(x^2-x+1)
∫(x -1/2) /(x^2-x+1) dx 凑微分, u = (x^2-x+1)
= (1/2)∫du / u = (1/2) lnu + C = (1/2) ln (x^2-x+1) + C
∫(1/2) /(x^2-x+1) dx = (1/2)∫dx / [(x-1/2)² + 3/4] 凑微分, v=(x-1/2)
= (1/2) ∫dv / (v² + 3/4) = (1/2) * (2 /√3) arctan(2v /√3) + C
= 1/√3 * arctan[(2x-1) /√3] + C
原式 = (1/2) ln (x^2-x+1) + (1 /√3) * arctan[(2x-1) /√3] + C

Answer 2

分母配方，换元t=x-1/2，则原式＝∫(t+1/2)/(t^2+3/4)dt＝∫t/(t^2+3/4)dt＋1/2×∫1/(t^2+3/4)dt。
后者套用公式∫dx/(x^2+a^2)＝1/a×arctan(x/a)+C得1/√3×arctan(2t/√3)＋C
前者化为1/2×∫(2t)/(t^2+3/4)dt＝1/2×∫1/(t^2+3/4)d(t^2+3/4)＝1/2×ln((t^2+3/4)+C
所以，原式＝1/2×ln((t^2+3/4)＋1/√3×arctan(2t/√3)＋C＝1/2×ln((x^2-x+1)＋1/√3×arctan((2x-1)/√3)＋C

Answer 3

∫x/(x^2-x+1)dx=∫x+1+1/xdx=x^2/2+x+lnx+c

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