- #1

jonc1258

- 5

- 0

**The problem statement**

Evaluate the indefinite integral

∫[itex]\frac{\sqrt{x}}{\sqrt{x}-3}[/itex]dx

**The attempt at a solution**

My first thought was to substitute u for √(x)-3, but then

*du*would equal [itex]\frac{1}{2\sqrt{x}}[/itex]dx, and there's no multiple of du in the integrand.

Next, I tried splitting up the problem like this:

∫[itex]\frac{1}{\sqrt{x}-3}[/itex]*[itex]\sqrt{x}[/itex]

When u is substituted for the [itex]\sqrt{x}[/itex],

*du*still equals[itex]\frac{1}{2\sqrt{x}}[/itex]dx , but the minus 3 in the denominator of the first factor prevents cleanly substituting a multiple of du for [itex]\frac{1}{\sqrt{x}-3}[/itex]dx.

Any ideas or hints?