Integration by u- substitution (involving natural logs)

[lLovePhysics]

Homework Statement

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

Homework Equations

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

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

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

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)

Why is this? Thanks in advance for your help!

 

[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.

 

[Dick]

The difference between your answer and the books answer is a CONSTANT, 1. You also have a C in your answer. What does that tell you?

 

[lLovePhysics]

The difference between your answer and the books answer is a CONSTANT, 1. You also have a C in your answer. What does that tell you?

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

 

[lLovePhysics]

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.

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

 

[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.

 

[lLovePhysics]

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.

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..

 

[Saladsamurai]

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

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

 

[Dick]

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..

Yes, they just sucked the 1 into the C.

 

[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;

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

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

 

[lLovePhysics]

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.

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