Integration question, u substitution

[jonc1258]

The problem statement
Evaluate the indefinite integral

∫$\frac{\sqrt{x}}{\sqrt{x}-3}$dx

The attempt at a solution
My first thought was to substitute u for √(x)-3, but then du would equal $\frac{1}{2\sqrt{x}}$dx, and there's no multiple of du in the integrand.

Next, I tried splitting up the problem like this:
∫$\frac{1}{\sqrt{x}-3}$*$\sqrt{x}$
When u is substituted for the $\sqrt{x}$, du still equals$\frac{1}{2\sqrt{x}}$dx , but the minus 3 in the denominator of the first factor prevents cleanly substituting a multiple of du for $\frac{1}{\sqrt{x}-3}$dx.

Any ideas or hints?

 

[Karnage1993]

Notice that if $u = \sqrt{x} - 3$, then $u + 3 = \sqrt{x}$.

Since $\displaystyle du = \frac{1}{2\sqrt{x}} dx, 2\sqrt{x}\ du = dx$

Does this make things easier?

 

[HallsofIvy]

$\sqrt{x}dx= (x/\sqrt{x})dx= (\sqrt(x)^2)(dx/\sqrt{x})= (u+3)^2du$

 

[jonc1258]

The denominator is kind of hard to see, it's sqrt(x)-3. Thanks for all the replies, I'll see if I can figure it out

 

[SammyS]

The problem statement
Evaluate the indefinite integral

∫$\frac{\sqrt{x}}{\sqrt{x}-3}$dx

The attempt at a solution
My first thought was to substitute u for √(x)-3, but then du would equal $\frac{1}{2\sqrt{x}}$dx, and there's no multiple of du in the integrand.

If $\displaystyle \ \ du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx\,,$

then $\displaystyle \ \ dx=2u\,du\ .$

Next, I tried splitting up the problem like this:
∫$\frac{1}{\sqrt{x}-3}$*$\sqrt{x}$
When u is substituted for the $\sqrt{x}$, du still equals$\frac{1}{2\sqrt{x}}$dx , but the minus 3 in the denominator of the first factor prevents cleanly substituting a multiple of du for $\frac{1}{\sqrt{x}-3}$dx.

Any ideas or hints?