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Anonymous
Anonymous asked in 教育與參考其他:教育 · 2 decades ago

數學問題...

兩題 integral 的問題

http://img.photobucket.com/albums/v336/mij_takuko/...

搞了半天也不知道對不對... 拜托幫幫忙吧~

算的方法跟解答~ 3q3q3q3q3q

Update:

z=arctan(x), sin(2z)=2x/(x^2+1)

我不懂....

b) 教授叫我們用這個方法...

http://www.math.ubc.ca/~israel/m103/e1prob_3.gif

Update 2:

ahahah~~ 不用啦~ 我懂啦~

2 Answers

Rating
  • 2 decades ago
    Favorite Answer

    (a) ∫1/(x^2+1)^2 dx

    令 x = tan(z),則 x^2 + 1 = sec^2(z),及 dx=sec^2(z) dz.

    ∫1/(x^2+1)^2 dx = ∫sec^2(z)/sec^4(z) dz = ∫cos(z)^2 dz = ∫(1+cos(2z))/2 dz = (z/2 + sin(2z)/4).

    因為z=arctan(x), sin(2z)=2x/(x^2+1),故

    ∫1/(x^2+1)^2 dx = (arctan(x) + x/(x^2+1))/2.

    (b) ∫(x^2-x)/(x^2+1)^2 dx

    (x^2-x)/(x^2+1)^2=(x^2+1-x-1)/(x^2+1)^2 = 1/(x^2+1) - x/(x^2+1)^2 - 1/(x^2+1)^2.

    第一項積分∫1/(x^2+1) dx = artctan(x),

    第二項積分∫x/(x^2+1)^2 dx = -0.5/(x^2+1),

    第三項積分∫1/(x^2+1)^2 dx = 0.5(arctan(x) + x/(x^2+1)) (由(a)可得).

    故∫(x^2-x)/(x^2+1)^2 dx = artctan(x) + 0.5/(x^2+1) - 0.5(arctan(x) + x/(x^2+1)) = 0.5arctan(x) + 0.5(1-x)/(x^2+1).

    補充:

    (a-1)因為x=tan(z),故z=arctan(x)=tan^{-1}(x),也就是tan()的反函數.(註:arctan(x)對x的微分 = 1/(x^2+1).)

    (a-2)因為x=tan(z),故sin(z)==x/√(x^2+1), cos(z)==1/√(x^2+1), (畫一個角度為z,底邊長1,垂直高長x的三角形可助了解),因此sin(2z)=2sin(z)cos(z)=2x/(x^2+1).

    (b) 教授叫我們用這個方法 --> 我的解答就是此法,將原式做部分分式分解,分母分別是(x^2+1)及(x^2+1)^2的,再個別去求各項的積分.

  • 2 decades ago

    請到這個網址抓答案吧

    因為都是數學式所以我把它弄成PDF檔

    http://homepage.ntu.edu.tw/~r93921082/a.pdf

    沒有寫的很完整

    有些變數變換的東西我沒有換回來

    Source(s): 自己
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