# U-substitution integral

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

Last edited:

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

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.

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

SammyS
Staff Emeritus