Re: [微積] 求dy/dx 跟積分

[Honor1984 (希望願望成真)]

※ 引述《zxcc79 (@!@)》之銘言：
: 1
: y=sin^2(cos(tan(x^2+1))) 求dy/dx


y' = 2sin(cos(tan(x^2+1))) * cos(cos(tan(x^2+1))) * (-1) sin(tan(x^2+1))

     * [sec(x^2+1)]^2 * 2x

: 2.∫1/(x(lnx)^3/2))dx

u = ln(x)

I = ∫du/u^(3/2) = -2u^(-1/2) + C = -2[ln(x)]^(-1/2) + C

: 3.∫(x-2)/(1+(x-1)^1/2)dx

u = 1 + sqrt(x-1)

I = ∫2(u-1)[(u-1)^2 - 1]/u  du

  = ∫2[(u-1)^3 - (u-1)]/u du

  = 2 ∫(u^3 - 3u^2 + 2u)/u du

  = 2 {(1/3)u^3 - (3/2)u^2 + 2u} + C

  剩下的就把u代回去
: 小第拜託幫忙了!!!

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◆ From: 128.220.159.5
→ zxcc79      : -2[ln(x)]^(-1/2) + C  可否化簡為ln(x)+c
→ zxcc79      : [sec(x^2+1)]^2 可否等於sec^2(x^2+1)