# Simple Integration (U-sub/LN)

1. Feb 4, 2008

### acurtiz

Hey everyone. One of the steps at the end of this problem is confusing to me. I'll point it out.

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

Setting u equal to $${1+ \sqrt{2x}}$$ and du equal to $$\frac {1}{\sqrt(2x)}$$ and taking the derivative, I get

$$\int \frac {sqrt(2x)}{u} du$$

The answer to the problem is apparently as follows -

$$\equiv \sqrt{2x} - ln|1+ \sqrt{2x}| + c$$

As far as I'm aware there is no step in between those last two. I'm not sure how it works and I'm 90% sure that I'm just missing something extremely obvious. I'd appreciate any help. Thank you!

Last edited: Feb 4, 2008
2. Feb 4, 2008

### cemar.

When you have a value under a square root your best bet is usually to set u = what is under that square root.
And i am very confused with your answer as it includes both variables u and x.
You must also remember to find a way to sub du into the value of dx.

3. Feb 4, 2008

### acurtiz

Sorry about that; hopefully my edited post is a little more clear. In my class we usually do set the value(s) under the square root as the u. However, in this particular problem, the directions are to set the entire denominator as u. I'm not sure how to do it even if I was to set u=2x. I'm new to the whole syntax thing. dx = sqrt(2x)du after going through everything.

4. Feb 4, 2008

### awvvu

$$u = 1 + \sqrt{2x} => \sqrt{2x} = u - 1$$

Does that help you? You need the integral to have the same variable.

5. Feb 5, 2008

### HallsofIvy

Staff Emeritus
After you have substituted "u" for "x", you cannot have an x remaining in the integral! I'm not sure how in the world you would integrate
$$\int \frac {sqrt(2x)}{u} du$$
with the x in there while differentiating with respec to u.

Fortunately, as awvvu said, since $u= 1+ \sqrt{2x}$, $\sqrt{2x}= u- 1$. You integral, after the substitution is really
$$\int \frac {u-1}{u} du = \int (1- \frac{1}{u}) du$$
That gives the answer you have.