Integral (極難之 6, 有興趣挑戰請進)

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參考資料：
My Maths knowledge

http://integrals.wolfram.com/index.jsp?expr=(tan+x)^(1%2F3)&random=false

上網用Integrator做左，不過好抱歉，我睇完都唔明佢點做出黎。

諗住一開始用by Parts，但做左又冇乜幫助，睇答案都知唔係用by Parts，
但係如果唔用by Parts，就更加唔知可以做D乜……

如果我研究到再補充啦！俾住個答案你參考先。

2008-11-26 08:29:15 補充：
Let u=tan^1/3 x, then x=tan^-1 u^3, dx=3u^2/(1+u^6) du
Thus I=∫tan^1/3 x dx=3∫u^3/(1+u^6) du

Consider u^6+1=(u^2+1)(u^2-√3 u+1)(u^2+√3 u+1)
so u^3/(1+u^6)=-1/3 u/(1+u^2)+1/6 u/(u^2-√3 u+1)+1/6 u/(u^2+√3 u+1)

超過左字數限制，只能做到呢度，但已經in得到。

tan^(1/3) x ??