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令 t = √[ (1+x)/x ]
t² = (1+x) / x = 1/x + 1
x = 1/(t² - 1)
dx = -2t / (t² - 1)² dt
∫ 1/x *√[(1+x)/x] dx
= ∫ (t² -1) * t* -2t / (t² - 1)² dt
= ∫ (-2t²) / (t² - 1) dt
= ∫ [ - 2 - 2/(t² - 1) ] dt
= ∫ [ -2 - 1/(t-1) + 1/(t+1) ] dt
= -2t - ln(t-1) + ln(t+1) +C
= - 2√[(1+x)/x] + ln [ (t+1)/(t - 1) ] + C
(t+1)/(t - 1)
= {√[(1+x)/x] +1} / { √[(1+x)/x] - 1}
= [ √(1+x) +√x ] / [√(1+x) - √x ]
= [√(1+x) + √x ]²
∴∫ 1/x *√[(1+x)/x] dx = 2ln [ √(1+x) + √x ] - 2√[ (1+x)/x ] + C
t² = (1+x) / x = 1/x + 1
x = 1/(t² - 1)
dx = -2t / (t² - 1)² dt
∫ 1/x *√[(1+x)/x] dx
= ∫ (t² -1) * t* -2t / (t² - 1)² dt
= ∫ (-2t²) / (t² - 1) dt
= ∫ [ - 2 - 2/(t² - 1) ] dt
= ∫ [ -2 - 1/(t-1) + 1/(t+1) ] dt
= -2t - ln(t-1) + ln(t+1) +C
= - 2√[(1+x)/x] + ln [ (t+1)/(t - 1) ] + C
(t+1)/(t - 1)
= {√[(1+x)/x] +1} / { √[(1+x)/x] - 1}
= [ √(1+x) +√x ] / [√(1+x) - √x ]
= [√(1+x) + √x ]²
∴∫ 1/x *√[(1+x)/x] dx = 2ln [ √(1+x) + √x ] - 2√[ (1+x)/x ] + C
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