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求(x^2+1)/(x^4+1)的不定积分
人气:470 ℃ 时间:2020-06-22 17:46:35
解答
积分:(x^2+1)/(x^4+1)dx =积分:(1+1/x^2)/(x^2+1/x^2)dx(上下同时除以x^2) =积分:d(x-1/x)/[(x-1/x)^2+(根号2)^2] =1/根号2*arctan[(x-1/x)/根号2]+C =1/根号2*arctan[(x^2-1)/(x根号2)]+C (C为常数)...
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