Solving Integrals with U-Substitution

[mmapcpro]

I'm trying to re-study the calculus I learned years ago to resume my degree this fall, so I don't have access to an instructor right now. I'm stuck on a problem that I can't figure out, and it's just bugging me.

Homework Statement

\int 1/(1 + \sqrt{2x}) dx

Homework Equations

I'm trying to use u-substitution, but I am not seeing it.

The Attempt at a Solution

I have thought to rationalize the denominator first, which would allow me to split it up into 2 integrals (1/1-2x) and (\sqrt{2x}/(1-2x) but I get stuck

 

[Dick]

Try u=1+sqrt(2x). What's dx in term of u?

 

[mmapcpro]

I set u = 1 + \sqrt{2x}, so then u-1 = \sqrt{2x}, so 2x = (u -1)^{2}

so then x = 1/2(u - 1)^{2}

then dx = (u - 1)du

so I could set it up as \int (u - 1)du / u

split that up into \int 1du - \int du/u

that leaves me with u - ln|u| + c

making it 1 + \sqrt{2x} - ln|1+\sqrt{2x}| + c

But the solution in the book reads: \sqrt{2x} - ln|1 + \sqrt{2x}| + c

I feel like an idiot...it's been about 10 years since I've done this.

 

[mmapcpro]

Does the +1 simply get absorbed into the constant, C?

 

[SammyS]

Does the +1 simply get absorbed into the constant, C?

Yes.