由衷感谢google大神 [不定积分][∫(x^n)/(1+(x^4)) dx] 寄件者：ㄚ肥呀 ([email protected]) 主旨：请问几题积分 ∫(x^n)/(1+(x^4)) dx ∫(x^n)/(1+(x^4)) dx 　　　　　　　　 n Ｘ ∫ --------------- ｄＸ 　　　　　　１＋(Ｘ^4) 1.n ＝　０...................台大考题....不是粉好做 2.n ＝　１...................easy 3.n ＝　２...................求救 4.n ＝　３...................so easy 5.n ＝　４...................回到 1 . 下面对不对我没看，请你参考看看 提供者 [email protected] ∫x^2/(1+(x^4)) dx = 因 1/(1+(x^4)) =1/((x^2+1)(x^2-√2 x+1)) + 1/((x^2+1)(x^2+√2 x+1)) => (x^2+1)/(1+(x^4))=1/(x^2-√2 x+1) + 1/(x^2+√2 x+1) =>∫x^2/(1+(x^4)) dx = ∫(x^2+1)/(1+(x^4)) dx -∫1/(1+(x^4)) dx =∫(1/(x^2-√2 x+1) dx + 1/(x^2+√2 x+1)) dx -∫1/(1+(x^4) dx) 上左式的前式的解为 ∫dx/(x^2-√2 x+1) =∫dx/((x-√2/2)^2 +1/2) (令 (x-√2/2) =√2/2tan y =>dx =√2/2.sec^2 y) =√2 y = √2 arctan (√2(x-√2/2)) ---(1) 上左式的二式的解为 ∫1/(x^2+√2 x+1)) dx =√2 arctan (√2(x+√2/2)) ---(2) 上左式的後式的解为 因 1/(1+(x^4)) =(1/2√2).1/(x(x^2-√2 x+1)) - 1/(x(x^2+√2 x+1)) =(1/2√2).((x +√2 )/(x^2+√2 x+1) - (x -√2 )/(x^2-√2 x+1)) => ∫ 1/(1+(x^4)) =(1/2√2).∫((x +√2 )/(x^2+√2 x+1) - (x -√2 )/(x^2 -√2 x+1)) dx 又 (1/2√2).∫(x +√2 )/(x^2+√2 x+1) dx =(1/2√2).∫x/(x^2+√2 x+1) dx +(1/2√2).∫√2/(x^2+√2 x+1) dx =-1/2 ln (1/((1+(x+√2/2)^2)^1/2) +(√2/2).arctan (√2(x+√2/2)) ----(3) (用 (2)式的方法可解之) 同理 (1/2√2).∫(x -√2 )/(x^2-√2 x+1) dx =-1/2 ln (1/((1+(x-√2/2)^2)^1/2) -(√2/2).arctan (√2(x-√2/2)) ※ 编辑: FATTY2108 来自: 218.184.96.125 (03/07 18:47)