Help finding an indefinite integral

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AbsValue13
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I am trying to find the following indefinite integral:

Homework Statement



∫[itex]\sqrt{}x[/itex]/(x-1)dx

Homework Equations



None

The Attempt at a Solution



I tried to use substitution but got nowhere. I set u=[itex]\sqrt{}x[/itex] so du=1/(2[itex]\sqrt{}x[/itex])dx. However from here on on I got stuck. I also tried using substitution this way u=x-1 so du=dx. However, this doesn't help since we get ∫[itex]\sqrt{}(u+1)[/itex]/(u)du.
 
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AbsValue13 said:
I am trying to find the following indefinite integral:

Homework Statement



∫[itex]\sqrt{}x[/itex]/(x-1)dx

Homework Equations



None

The Attempt at a Solution



I tried to use substitution but got nowhere. I set u=[itex]\sqrt{}x[/itex] so du=1/(2[itex]\sqrt{}x[/itex])dx. However from here on on I got stuck. I also tried using substitution this way u=x-1 so du=dx. However, this doesn't help since we get ∫[itex]\sqrt{}(u+1)[/itex]/(u)du.

The substitution u = √x will work.
So u2 = x => 2udu = dx.

You'll get an improper fraction that you can simplify using polynomial division.

LaTeX tip: Put the quantity that's inside the radical inside the braces {}. IOW, \sqrt{u + 1}.
 
AbsValue13 said:
I am trying to find the following indefinite integral:

Homework Statement



∫[itex]\sqrt{}x[/itex]/(x-1)dx

Homework Equations



None

The Attempt at a Solution



I tried to use substitution but got nowhere. I set u=[itex]\sqrt{}x[/itex] so du=1/(2[itex]\sqrt{}x[/itex])dx. However from here on on I got stuck. I also tried using substitution this way u=x-1 so du=dx. However, this doesn't help since we get ∫[itex]\sqrt{}(u+1)[/itex]/(u)du.
Or, you can use ASCII and write sqrt(x/(x-1)) or sqrt(x)/(x-1).

I cannot figure out whether your integrand is
[tex]\sqrt{\frac{x}{x-1}} \text{ or } \frac{\sqrt{x}}{x-1}[/tex]
If you mean the first one, use "[t e x ] \sqrt{ \frac{x}{x-1} } [/t e x ]" (no spaces); if you mean the second one, use "[t e x ] \frac{ \sqrt{x} }{x-1} [/ t e x]" (no spaces).