作者yuyumagic424 (五月飞雪)
看板trans_math
标题Re: [积分] 请问一下一题不定积分
时间Thu Nov 3 05:06:38 2005
※ 引述《chenshinwei (chen)》之铭言：
: x^(1/3)
: ∫------------- dx
: 1+x^(2/3)
: 谢谢各位的解答
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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