Can someone please tell me how my book did this integration

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Homework Help Overview

The discussion revolves around the integration of a differential equation involving the expression (-1 + \frac{2}{1+u}) du + \frac{dx}{x} = 0. Participants are examining the transition from this expression to the form -u + 2ln(1+u) + ln(x) = ln(c), questioning the reasoning behind the integration process and the appearance of ln(c) as a constant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants express confusion regarding the integration of zero and the nature of constants in integration. There are attempts to clarify the relationship between constants and logarithmic forms, with some questioning the necessity of using ln(c) instead of a simpler constant notation. Others seek to understand the specific steps leading to the derived expressions.

Discussion Status

The discussion is active, with participants sharing insights and clarifications about the integration process. Some have acknowledged understanding the equivalence of constants in different forms, while others continue to seek clarity on specific aspects of the integration and notation used in the book.

Contextual Notes

Participants note the late hour of discussion and the potential for confusion due to fatigue. There is also mention of an initial error in the transcription of the differential equation, which has since been corrected.

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Homework Statement


Attached is a picture..

the book went from
(-1 + \frac {2}{1+u} ) du + \frac {dx}{x} = 0
Edited by moderator.

to

-u + 2ln(1+u) + ln(x) = ln(c)

Homework Equations

The Attempt at a Solution



How in the world could the integral of 0 be ln(c)?

Maybe because it's 5 am and I'm up late and I'm hallucinating, but I always thought the integral of 0 is c, just a constant.

taking the integral of

(-1 \frac {2}{1+u} ) du + \frac {dx}{x} = 0

I get

(-u + 2 ln(u+1) + c ) + ln(x) + c_2 = c_3

-u +2ln(u+1) + ln(x) = c_4
 

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Rijad Hadzic said:
How in the world could the integral of 0 be ln(c)?
It is an integration constant ...

Edit: And you seem to be getting exactly the same so I do not see what the problem is ...

Edit 2: Obviously, the logarithm of a constant is a constant ...
 
Orodruin said:
It is an integration constant ...

Edit: And you seem to be getting exactly the same so I do not see what the problem is ...

Edit 2: Obviously, the logarithm of a constant is a constant ...

I understand that but why not just write K or whatever, why include ln? Like where did ln come from? They later use properties of logarithms to write

\frac yx = ln(\frac {(x+y)^2}{cx}) so I need to know where the ln(c) came from..
 
I mean I guess it makes sense since ln(c) is just a constant... I guess I've just never seen anything like this before so its new to me..
 
You can use either ##c## or ##\ln(c)##, it makes absolutely no difference - both are arbitrary constants. The reason they chose ##\ln(c)## is that they could take it inside the logarithm when simplifying their expression. Again, the expressions are completely equivalent.
 
Orodruin said:
You can use either ##c## or ##\ln(c)##, it makes absolutely no difference - both are arbitrary constants. The reason they chose ##\ln(c)## is that they could take it inside the logarithm when simplifying their expression. Again, the expressions are completely equivalent.

Gotcha. I mean it makes sense now that I think about it... thank you brother.
 
Rijad Hadzic said:

Homework Statement


Attached is a picture..

the book went from
(-1 \frac {2}{1+u} ) du + \frac {dx}{x} = 0

to

-u + 2ln(1+u) + ln(x) = ln(c)

Homework Equations

The Attempt at a Solution



How in the world could the integral of 0 be ln(c)?

Maybe because it's 5 am and I'm up late and I'm hallucinating, but I always thought the integral of 0 is c, just a constant.

taking the integral of

(-1 \frac {2}{1+u} ) du + \frac {dx}{x} = 0

I get

(-u + 2 ln(u+1) + c ) + ln(x) + c_2 = c_3

-u +2ln(u+1) + ln(x) = c_4

The integral of
$$\left(-1 \frac{2}{1+u} \right) du + \frac{1}{dx} = 0$$
is
$$ - 2 \ln(1+u) + \ln(x) = \text{constant}$$
How did you get ##-u + 2 \ln(1+u)?##
 
Ray Vickson said:
The integral of
$$\left(-1 \frac{2}{1+u} \right) du + \frac{1}{dx} = 0$$
is
$$ - 2 \ln(1+u) + \ln(x) = \text{constant}$$
How did you get ##-u + 2 \ln(1+u)?##
If you look at the image, he just copied it wrong. The expression is not
$$\left(-1 \frac{2}{1+u} \right) du + \frac{1}{dx} = 0$$
it is
$$\left(-1 + \frac{2}{1+u} \right) du + \frac{1}{dx} = 0$$
 
It's fixed now in post #1.
 

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