ä»¤æ ¹å·x-1=t
x=t²ï¼1
dx=2tdt
åå¼=â« (t²ï¼1)/t · 2tdt
=2â«(t²ï¼1)dt
=2/3t³ï¼2tï¼c
=2/3 ï¼æ ¹å·x-1ï¼³ï¼2(æ ¹å·x-1)+c
|
|
|
-===+=====_/(T)\_=====+===-
| |/.\| |
`-|\_/|-'
x=t²ï¼1
dx=2tdt
åå¼=â« (t²ï¼1)/t · 2tdt
=2â«(t²ï¼1)dt
=2/3t³ï¼2tï¼c
=2/3 ï¼æ ¹å·x-1ï¼³ï¼2(æ ¹å·x-1)+c
|
|
|
-===+=====_/(T)\_=====+===-
| |/.\| |
`-|\_/|-'
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第1个回答 2013-05-27
解：令 √x-1=t 则x=(t+1)² dx=2(t+1)dt
∫x/(√x-1)dx=∫(t+1)²/t*2(t+1)dt
=2∫(t³+3t+3+1/t)dt
=2t³/3+3t² +6t+2lnt+C
∫x/(√x-1)dx=∫(t+1)²/t*2(t+1)dt
=2∫(t³+3t+3+1/t)dt
=2t³/3+3t² +6t+2lnt+C
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