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# 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

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

我不懂....

b) 教授叫我們用這個方法...

ahahah~~ 不用啦~ 我懂啦~

### 2 Answers

- 2 decades agoFavorite 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 agoSource(s): 自己