求不定积分:1/(x+sqrt(x^2-x+1))
人气:402 ℃ 时间:2020-04-28 18:50:50
解答
1/(x+sqrt(x^2-x+1))=[sqrt(x^2-x+1)-x]\(1-x)sqrt(x^2-x+1)\(1-x)dx=-sqrt(t^2-t+1)\tdtt=1-x-sqrt(t^2-t+1)\t=1\2sqrt(t^2-t+1)-(t-1\2)\sqrt(t^2-t+1)-1\tsqrt(t^2-t+1)u=1\t-1\tsqrt(t^2-t+1)dt=sgnu\sqrt(u...
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