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数学
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求不定积分dx/根号(1+x-x^2)请老师详细一点谢谢
人气:294 ℃ 时间:2020-01-29 11:49:15
解答
原式=∫dx/√-(x^2-x+1/4-1/4-1)
=∫dx/√[5/4-(x-1/2)^2]
=∫dx/√[(√5/2)^2-(x-1/2)^2]
=arcsin[(x-1/2)/(√5/2)]+C
=arcsin[(2x-1)/√5]+C.
利用三角代换证明公式∫dx/√(a^2-x^2)=arcsin(x/a)+C.
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