請問一題不定積分:
http://www.wretch.cc/album/show.php?i=flank0001&b=22&f=1842577104&p=4
謝謝...
請問一題微積分的問題
2013-08-12 2:27 am
回答 (2)
2013-08-12 2:50 am
✔ 最佳答案
Dear Flank, please read:圖片參考：http://imgcld.yimg.com/8/n/HA00430218/o/20130811185010.jpg
2013-08-11 18:51:30 補充：
Please delete the final "=" sign.
2013-08-12 3:17 am
∫dx/(x^4+4)=∫(dx/x^2 )/(x^2+4/x^2 )=1/4 [∫(1+2/x^2 )dx/((x-2/x)^2+4)-∫(1-2/x^2 )dx/((x+2/x)^2-4)]
=1/4 [∫d(x-2/x)/((x-2/x)^2+4)-1/4 ∫(1/(x+2/x-2)-1/(x+2/x+2))d(x+2/x) ]
=1/8 tan^(-1)〖(x-2/x)/2〗-1/16 ln|(x+2/x-2)/(x+2/x+2)|+C
=1/8 tan^(-1)〖(x^2-2)/2x〗+1/16 ln((x^2+2x+2)/(x^2-2x+2))+C
=1/4 [∫d(x-2/x)/((x-2/x)^2+4)-1/4 ∫(1/(x+2/x-2)-1/(x+2/x+2))d(x+2/x) ]
=1/8 tan^(-1)〖(x-2/x)/2〗-1/16 ln|(x+2/x-2)/(x+2/x+2)|+C
=1/8 tan^(-1)〖(x^2-2)/2x〗+1/16 ln((x^2+2x+2)/(x^2-2x+2))+C
收錄日期: 2021-04-11 19:59:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130811000016KK03173