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求不定积分arctan(1/x)/(1+x2)dx
人气:400 ℃ 时间:2020-03-23 02:31:01
解答
令1/x = t
则原式=
∫arctant/(1 + 1/t²) * (-1/t²)dt
=∫-arctant/(1+t²) dt
=∫-arctant darctant
=-1/2 arctan²t + C
=-1/2 arctan(1/x²) + C
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