※ 引述《chribaby (过去˙现在˙未来)》之铭言： : http://www.lib.ntu.edu.tw/exam/graduate/91/91049.htm : 2. 设f(x)=x^nln(x)，则f^(n+1)(x)=? f^(1)(x) = nx^(n-1)ln(x)+x^n*(1/x) = nx^(n-1)ln(x)+x^(n-1) = n*[ln(x)+(1/n)]x^(n-1) f^(2)(x) = n(n-1)*[ln(x)+(1/n)]x^(n-2)+nx^(n-2) = n(n-1)*[ln(x)+(1/n) + 1/(n-1)]x^(n-2) f^(3)(x) = n(n-1)(n-2)*[ln(x)+(1/n) + 1/(n-1)]x^(n-3)+n(n-1)x^(n-3) = n(n-1)(n-2)*[ln(x) + 1/n + 1/(n-1) + 1/(n-2)]x^(n-3) . . . f^(n)(x) = (n!/1!)*[ln(x) + 1/n + 1/(n-1) +...+ 1/2]x f^(n+1)(x) = (n!)*[ln(x) + 1/n + 1/(n-1) +...+ 1/2]+(n!/1!) = (n!)*[ln(x) + 1/n + 1/(n-1) +...+ 1/1] : 这题感觉要用泰勒(是这样吗?) : 但却不知道要怎麽下手 : 拜托帮帮我~~ : 4. 定积分 S(pi/4~pi/2)1/(sinx)^4 dx π/2 π/2 ∫ 1/(sinx)^4 dx = ∫ (cscx)^2*(cscx)^2 dx π/4 π/4 π/2 = -∫ (cscx)^2 d(cotx) π/4 π/2 = -∫ [1+(cotx)^2] d(cotx) π/4 |π/2 = -[(cotx)+(cotx)^3/3]|π/4 = -[0+0/3]+[1+1/3] = 4/3 --

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