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求不定积分:根号下[(1-x) / (1+x)] dx
人气:425 ℃ 时间:2019-12-05 10:41:34
解答
∫√[(1-x)/(1+x)]dx
=∫(1-x)/√(1-x^2)dx
=∫1/√(1-x^2)-∫x/√(1-x^2)dx
=arcsinx+1/2∫(1-x^2)^(-1/2)d(1-x^2)
=arcsinx+√(1-x^2)+c
c是常数
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