integration

留上用的方法只是simple substitution, 並不是 by part, 要小心~

Integration by part:

∫x/(4+x^2)dx

Theorem:
∫u dv/dx dx = uv - ∫v du/dx dx ...(*)

Let u = x
Let dv/dx = 1/(4+x^2)

du/dx = dx/dx = 1
v = ∫dv/dx dx = ∫1/(4+x^2)dx=1/2 arctan(x/2)

Substituting back into (*)
∫u dv/dx dx
= (x) (1/2 arctan(x/2)) - ∫(1/2 arctan(x/2)) (1) dx
= x/2(arctan(x/2)) - 1/2 (x*arctan(x/2)-ln(x^2+4))
= x/2(arctan(x/2)) - x/2(arctan(x/2)) + 1/2(ln(x^2+4))
= 1/2(ln(x^2+4))

2011-04-24 21:06:02 補充：
基本上 1/2 ∫(arctan(x/2)) (1) dx = 1/2 (x*arctan(x/2)-ln(x^2+4))
但詳盡解釋這個要比較長時間+用圖, 看你這只是5分題, 你還是自己看看數學書學一下如何Integrate arctan 這種function吧。

2011-04-25 18:36:47 補充：
我覺得是因為問者在做integration by part 的exercise吧?
1+1 = 2 , 2*1 也是=2, 如果你不用一下其他方法, 如何學新東西呢~

一定要用integration by part? 用integration by substitution似乎快d!

∫x(4+x^2)^-1 dx let u=4+x^2
=1/2 ∫ u^-1 du du=2x dx
=1/2 ln(4+x^2) +C