int[x^0.5/(1+x^2)]dx
x:0~infinity
積分積分積分積分積分積分積分
2009-12-27 10:45 am
回答 (1)
2009-12-27 9:11 pm
✔ 最佳答案
令 t^2=x, 則原式=A=∫[0~∞] 2t^2/(1+t^4) dt再令 t= 1/x, 則A=∫[0~∞] 2/(x^4+1) dx = B
故原式=(A+B)/2=∫[0~∞] (1+x^2)/(1+x^4) dx
(同除以 x^2) =∫[0~∞] d(x-1/x) / [ (x-1/x)^2 +2 ]
= 1/√2 * arctan [(x-1/x)/√2] 代 x=0~∞
=π/√2
收錄日期: 2021-05-04 00:46:09
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091227000010KK01011