# Integration by u- substitution (involving natural logs)

1. Dec 23, 2007

### lLovePhysics

1. The problem statement, all variables and given/known data
$$\int \frac{1}{1+\sqrt{2x}}dx$$

2. Relevant Equations

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

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

$$dx=(u-1)du$$

3. The attempt at a solution

I was able to get it down to:

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

$$= u-\ln{lul}}+C$$

$$= 1+\sqrt{2x}-\ln{l1+\sqrt{2x}l}+C$$

However, my book says that the solution to the integral is:

$$\sqrt{2x}-\ln{l{1+\sqrt{2x}l}+C$$ (Without the 1 in front)

Last edited: Dec 23, 2007
2. Dec 23, 2007

### rocomath

well i got

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

be careful with your re-substitution ... b/c you're "u-sub" was $$1+\sqrt{2x}$$ so you need parenthesis.

3. Dec 23, 2007

### Dick

4. Dec 23, 2007

### lLovePhysics

It means that C=-1 but how do you find that out? What points are there to substitute to obtain C?

5. Dec 23, 2007

### lLovePhysics

How do you put absolute value signs around an expression? I've tried typing \abs{expression} but it doesn't work.

6. Dec 23, 2007

### Dick

If you click on rocophysics answer, you'll see the tex. He just used a vertical bar. The point to the C thing is that an indefinite integral is only defined up to an arbitrary constant. You can absorb the 1 into the C (as the book answer did). If you take the derivative of both answers you'll get the same thing. You are both right.

7. Dec 23, 2007

### lLovePhysics

Ohh.. so my book just sucked in all of the constants and made it an arbitrary constant C to represent all constants (including +1)?

How do you put a vertical bar? I've tried inserting the lower case L but to no avail..

8. Dec 23, 2007

It should be a key on your keyboard above enter button. SHIFT+\ = |

9. Dec 24, 2007

### Dick

Yes, they just sucked the 1 into the C.

10. Dec 25, 2007

### Gib Z

It's probably better to think of $$f(x) + C$$ as representing a family of functions, rather than a function with some constant. That way;

instead becomes a redundant question, since f(x) + 1 + C has exactly the same members as f(x) + C.

11. Dec 25, 2007

### lLovePhysics

Yeah, I think it's better to think of it as a family of functions too. Thanks