Re: [積分] 台大82甲
※ 引述《feathersss (不定)》之銘言：
: 請問一下
: 1
: ___________________
: x^(1/3) (1+x)^(5/3)
: 對x做積分該怎麼做呢?
I=∫[x^(-1/3)][(1+x)^(-5/3)]dx
令 u = [(1+x)/x]^(1/3)
所以 (1+x)/x = u^3 因此 [-1/(x^2)]dx = 3u^2du
I = ∫[(1+x)/x]^(1/3)*{(x^2)/[(1+x)^2*x^2]}dx
= ∫[(1+x)/x]^(1/3)*[x/(1+x)]^2*(1/x^2)dx
= ∫u*u^(-6)(-3u^2)du
=-3∫u^(-3)du
=(3/2)*u^(-2) + c
=(3/2)*[(1+x)/x]^(-2/3) + c
=(3/2)*[x/(1+x)]^(2/3) + c
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