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Integral of sqrt((x)/(x-1))

  • Thread starter Siune
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  • #1
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Homework Statement


[itex]\int[/itex]([itex]\sqrt{x}/\sqrt{x-1}[/itex] )dx.


Homework Equations


-


The Attempt at a Solution



It should be doable with substitution or/and with partial intergral. I just don't figure out what to substitute. I have tried with u = √(x-1), u = √(x), and with partial integral formula:

∫u*v´ = u*v - ∫v * u´

Any tips?

Thanks for any help
-Siune
 

Answers and Replies

  • #2
SammyS
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Homework Statement


[itex]\int[/itex]([itex]\sqrt{x}/\sqrt{x-1}[/itex] )dx.

Homework Equations



The Attempt at a Solution



It should be doable with substitution or/and with partial intergral. I just don't figure out what to substitute. I have tried with u = √(x-1), u = √(x), and with partial integral formula:

∫u*v´ = u*v - ∫v * u´

Any tips?

Thanks for any help
-Siune
Hello Siune. Welcome to PF .

Try the substitution, u = x-1 .
 
  • #3
HallsofIvy
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I think much simpler is to let [itex]u= \sqrt{x}[/itex]. Now what are dx and x- 1 in terms of u and du?
 
  • #4
SammyS
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Hello Siune. Welcome to PF .

Try the substitution, u = x-1 .
The result of this is no better than the original.
 
  • #5
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don't use partial integral, first multiply the integrand ∫(√x/√(x−1))dx to √x/√x..

Moderator note: I removed the subsequent work shown. Please let the OP try to work out the problem on his or her own.
 
Last edited by a moderator:
  • #6
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^

Adding that extra sqrt(x) was clever. I seem to understand and accept with everything, but there is the part

"divide 2x-1 to x"?

U mean I calculate u = 2x-1 [itex]\Leftrightarrow[/itex] x = (1/2)(u+1)?
which is then x dx = (1/2)(u+1) du?


I'm sorry I might seem like totally idiot, but until university, sign (dx/du) was totally unknown to me so I'm not familiar with it and don't know how it exactly behaves.

To HallsOfIvy, thanks for the tip.
 

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