How do I solve the integral of sqrt(x)/(sqrt(x-1))?

[Siune]

Homework Statement

$\int$($\sqrt{x}/\sqrt{x-1}$ )dx.

Homework Equations

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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

 

[SammyS]

Homework Statement

$\int$($\sqrt{x}/\sqrt{x-1}$ )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 .

 

[HallsofIvy]

I think much simpler is to let $u= \sqrt{x}$. Now what are dx and x- 1 in terms of u and du?

 

[SammyS]

Hello Siune. Welcome to PF .

Try the substitution, u = x-1 .

The result of this is no better than the original.

 

[engphy2]

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.

 

[Siune]

^

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 $\Leftrightarrow$ 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.