Re: [積分] 還是一題不定積分
※ x^4 + 1
: ∫------------dx
: x^6 + 1
: 引述《Karter (偽Carter)》之銘言：
: 先謝謝了 <(_ _)>
x^4 + 1
I= ∫------------dx
x^6 + 1
let x=tanA
do the integration, you may obtain
(tanA^2 + 1)tanA^2
I=A+∫-------------------dA where tanA^2 means (tanA)^2
tanA^6 + 1
Then let tanA^2=u, du du
dA=---------------=--------------
2tanA secA^2 2u^1/2 (u+1)
(tanA^2 + 1)tanA^2 u^1/2 du du^2/3
Thus,∫-------------------dA=∫-------------------=1/3∫-------------
tanA^6 + 1 2(u^3+1) u^3+1
=1/3arctan(u^2/3)
=1/3arctan(tanA^3)
Now, I=arctan(x)+arctan(x^3)
or we may write it as I=2/3arctan(x)-1/3arctan[x/(x^2-1)]
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.112.251.241
推
140.127.179.227 07/09, , 1F
140.127.179.227 07/09, 1F
推
61.228.25.202 07/09, , 2F
61.228.25.202 07/09, 2F
推
218.168.194.15 07/10, , 3F
218.168.194.15 07/10, 3F
討論串 (同標題文章)
完整討論串 (本文為第 2 之 3 篇)：
積分
1
1