Can someone please tell me how my book did this integration

[Rijad Hadzic]

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

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

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

 

[Orodruin]

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 ...

 

[Rijad Hadzic]

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..

 

[Rijad Hadzic]

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..

 

[Orodruin]

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.

 

[Rijad Hadzic]

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.

 

[Ray Vickson]

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 $-u + 2 \ln(1+u)?$

 

[Orodruin]

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
$$\left(-1 + \frac{2}{1+u} \right) du + \frac{1}{dx} = 0$$

 

[Mark44]

It's fixed now in post #1.