Re: [積分] 請問一下一題不定積分
※ 引述《chenshinwei (chen)》之銘言：
: x^(1/3)
: ∫------------- dx
: 1+x^(2/3)
: 謝謝各位的解答
Let u = x^(2/3) + 1
then du = (2/3) x^(-1/3) dx
x^(1/3) = x^(-1/3) ． x^(2/3)
x^(2/3) = u - 1
x^(1/3)
==> ∫------------- dx
1+x^(2/3)
(3/2) (u-1)
= ∫------------- du
u
= (3/2) ∫(1-1/u) du
.
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