一題積分的問題...

2005-07-30 7:20 pm
請問 ∫1/(1+ x^4) dx 怎麼算啊!

回答 (4)

2005-07-30 10:02 pm
✔ 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!
2005-08-01 2:57 am
樓上給的計算方法、答案皆是錯的,完全錯誤!
我用 Maple 計算軟體作積分,以下是答案:

1/8*2^(1/2)*ln((x^2+x*2^(1/2)+1)/(x^2-x*2^(1/2)+1))+1/4*2^(1/2)*arctan(x*2^(1/2)+1)+1/4*2^(1/2)*arctan(x*2^(1/2)-1)

好複雜呀!

2005-07-31 18:58:57 補充:
請用三角函數計算積分,如:設 x = tan(x)…等!
2005-07-30 10:35 pm
這個題目的積分後函數,好像是屬於
"非初等函數",(非有限項).還有請問
加冰不加糖的解法好像少了蛇麼?

令u = x2, du= 2x dx 才對呀

少了 2x 這項
2005-07-30 10:01 pm
Sorry...我承認我不會,也懶再算........................
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