✔ 最佳答案
本題用到因式分解、部分分式的化法(此兩項請自行演算)及兩個積分公式:
(1)∫ du/u = ln|u|+c
(2)∫du/(u^2 + a^2) = 1/a(arctan u/a)+c
解: 1 /(x^4 + 1)= 1 /(x^2+√2x+ 1)(x^2-√2x+ 1)
=(√2x/4 + 1/2)/(x^2+√2x+1)- [(√2x/4 - 1/2 )/(x^2-√2x+1)]
=(√2/8)(2x+√2)/(x^2+√2x+1) + (1/4)/(x^2+√2x+1)
-(√2/8)(2x-√2)/(x^2-√2x+1) + (1/4)/(x^2-√2x+1)
所以 ∫dx/(x^4 + 1)
=∫[(√2/8)(2x+√2)/(x^2+√2x+1)]dx+∫[(1/4)/(x^2+√2x+1)]dx-∫[(√2/8)(2x-√2)/(x^2-√2x+1)]dx+∫[(1/4)/(x^2-√2x+1)]dx
=(√2/8)∫d(x^2+√2x+1)/(x^2+√2x+1)-(√2/8)∫d(x^2-√2x+1)/(x^2-√2x+1)+∫[(1/4)/(x+√2/2)^2 +(√2/2)^2]dx+∫[(1/4)/(x-√2/2)^2 +(√2/2)^2]dx
=(√2/8)ln|x^2+√2x+1|-(√2/8)ln|x^2-√2x+1|
+(1/4).(1/√2/2)arctan (x+√2/2)/√2/2)
+(1/4).(1/√2/2)arctan (x-√2/2)/√2/2)+c
=(√2/8)ln|x^2+√2x+1|-(√2/8)ln|x^2-√2x+1|
+(√2/4)arctan(√2x+1)+(√2/4)arctan(√2x-1)+c
2005-07-31 21:12:37 補充:
對或錯?將答案微分看會不會等於原被積函數就知道了!