# Integral (1/(1+sqrt(2x))) dx

1. Jan 7, 2008

### johnsonandrew

1. The problem statement, all variables and given/known data
Find the indefinite integral.
∫ (1/(1+sqrt(2x))) dx

2. Relevant equations
∫ 1/u du = ln |u| + C

3. The attempt at a solution
I tried a couple 'u' substitutions, which didn't work out. I also tried rationalizing the denominator, but that didn't help. No one I've talked to knows how to do this one...

2. Jan 7, 2008

### rocomath

Well from rationalizing we get ...

$$\int\left(\frac{1}{1-2x}-\frac{\sqrt{2x}}{1-2x}}\right)dx$$

So from here, the left is easy and now we work only with the right

$$-\int\frac{\sqrt{2x}}{1-2x}dx$$

$$u=\sqrt{2x}\rightarrow u^2=2x$$

$$u^2=2x \leftrightarrow udu=dx$$

Last edited: Jan 7, 2008
3. Jan 7, 2008

### johnsonandrew

following you so far

4. Jan 7, 2008

### rocomath

After substituting, we get ...

$$\int\frac{-u^2}{1-u^2}du$$

Then add $$\pm 1$$ to the numerator so that you can split it into 2.

$$\int\frac{(1-u^2)-1}{1-u^2}du$$

Last edited: Jan 7, 2008
5. Jan 7, 2008

### johnsonandrew

I don't understand:
$$u^2=2x \leftrightarrow udu=dx$$

6. Jan 7, 2008

### rocomath

I made my initial u-sub then I manipulated my u-sub by squaring both sides and then I took it's derivative.

$$u=\sqrt 2x$$ ONLY for the numerator

Manipulating my u-sub by squaring both sides so that I can substitute for my denominator.

$$u^2=2x$$

Taking the derivative of my manipulating u-sub

$$2udu=2dx \rightarrow udu=dx$$

7. Jan 7, 2008

### johnsonandrew

Ohhh okay. Thank you!

8. Jan 7, 2008

Anytime.

9. Jan 7, 2008

### rocomath

Actually, rationalizing isn't even a good idea. You can apply the same methods I did with the u-sub w/o rationalizing.