Re: [微積] 積分部分分式

[wayne2011 (乃瑜讓我想流鼻血)]

※ 引述《rita42027 (CHIEN)》之銘言：
: http://i.imgur.com/x46zMlZ.jpg
: 請問(c)怎麼寫？

f(x)=(2x+1)/[(x+1)(x-1)^2]

= A/(x+1) + B/(x-1) + C/(x-1)^2


A=(x+1)f(x)│    = -1/4
            x=-1

B=[(x-1)^2*f(x)]'│   = [1/(x+1)^2]│   = 1/4
                  x=1               x=1

C=(x-1)^2*f(x)│   = 3/2
               x=1

原式

=∫(2x+1)/[(x+1)(x-1)^2] dx

=∫ [(-1/4)/(x+1) + (1/4)/(x-1) + (3/2)/(x-1)^2] dx

= -(1/4)ln│x+1│+ (1/4)ln│x-1│- (3/2)[1/(x-1)] + c

= (1/4)ln│(x-1)/(x+1)│- 3/[2(x-1)] + c









--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 122.100.91.75
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1451013465.A.139.html
※ 編輯: wayne2011 (122.100.91.75), 12/25/2015 11:18:54