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Integral(SOME hard)

Integral(SOME hard)

Thx very much, the question is really complicated




[ 本帖最後由 小白無對手 於 2010-11-13 08:39 PM 編輯 ]
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【DSE Maths】中大教育文憑 × 數學系畢業◆助你奪*升Lv~
首次以課題分班,10月開班招生,快d黎睇下~

導師簡介:按此進入

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You may do the following substitution
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integral 1/(-1+4 e^(2 x)) dx

For the integrand 1/(4 e^(2 x)-1), substitute u = 2 x and  du = 2 dx:
= 1/2 integral 1/(4 e^u-1) du

For the integrand 1/(4 e^u-1), substitute s = e^u and  ds = e^u du:
= 1/2 integral 1/(s (4 s-1)) ds

For the integrand 1/(s (4 s-1)), use partial fractions:
= 1/2 integral (4/(4 s-1)-1/s) ds

Integrate the sum term by term and factor out constants:
= 2 integral 1/(4 s-1) ds-1/2 integral 1/s ds

For the integrand 1/(4 s-1), substitute p = 4 s-1 and  dp = 4 ds:
= 1/2 integral 1/p dp-1/2 integral 1/s ds

The integral of 1/p is ln(p):
= (ln(p))/2-1/2 integral 1/s ds
The integral of 1/s is ln(s):
= (ln(p))/2-(ln(s))/2+constant
Substitute back for p = 4 s-1:
= 1/2 ln(4 s-1)-(ln(s))/2+constant
Substitute back for s = e^u:
= 1/2 ln(4 e^u-1)-(ln(e^u))/2+constant
Substitute back for u = 2 x:
= 1/2 ln(4 e^(2 x)-1)-1/2 ln(e^(2 x))+constant
Factor the answer a different way:
= 1/2 (ln(4 e^(2 x)-1)-ln(e^(2 x)))+constant
An alternative form of the integral is:
= 1/2 ln(4-e^(-2 x))+constant

when x=ln 5
(1/2) ln(4-e^(-2ln5))+C
=(1/2) ln (4-e^(ln(1/25))+C
=(1/2) ln (4-0.04)+C
=(1/2) ln3.96+C
=(1/2) ln(99/25)+C

when x=ln 3
(1/2) ln(4-e^(-2ln 3))+C
=(1/2) ln (4-1/9)+C
=(1/2) ln (35/9)+C

the answer:
=(1/2) ln (99/25)-(1/2) (ln 35/9)
=(1/2) ln[(99/25)x)9/35)]
=(1/2) ln(891/875)

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搞咩???

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