Re: [積分] 令人猜不透的瑕積分
※ 引述《announcer (announcer)》之銘言：
: ∞ 1
: ∫ --------------dx
: 1 x + x^(√2)
: 解答是令 u = x^(√2 - 1)
: 接下來就可以順利解出答案為(√2 +1)ln2
: 但重點是...怎麼思考才會想到令 u = x^(√2 - 1)呢?!
: 煩請大家指教 謝謝!
∞ 1 l i m b dx
∫ -----------------dx = ∫ --------------------
1 x + x^(√2) b->∞ 1 x( 1 + x^(√2 - 1))
1 lnu
令u = x^(√2 - 1) => dx = ---------- e^[-----------]du
u(√2 - 1) (√2 - 1)
∞ 1 l i m ∞ dx
∫ --------------dx = ∫ --------------------
1 x + x^(√2) b->∞ 1 x( 1 + x^(√2 - 1))
l i m c e^[lnu/(√2 - 1)]
= ∫ ------------------------------------------ du
c->∞ 1 e^[lnu/(√2 - 1)] ( 1 + u ) u (√2 - 1)
l i m c du
= ∫ ---------------------
c->∞ 1 ( 1 + u ) u (√2 - 1)
1 l i m u u=c
= ----------- ln (---------)|
(√2 - 1) c->∞ 1 + u u=1
ln(2)
= -----------
(√2 - 1)
^^^^^^^^^^^^^^^^^^ <(^ ﹏ ^)/
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◆ From: 61.229.160.5
※ 編輯: GayerDior 來自: 61.229.160.5 (03/30 00:38)
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