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※ 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)]
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