Where did this 1/4*ln(C) term come from when integrating?

[dchau503]

Homework Statement

The givenequation is this: <br /> \frac{1}{4} \frac {du}{(2-u)} \ + \ \frac{1}{4} \frac{du}{(2+u)} \ = \frac{dx}{x}
My book says that when integrated, the above equation becomes <br /> \frac{-1}{4} \ln (2-u) \ + \ \frac{1}{4} \ln (2+u) \ = \ln (x) + \frac{1}{4} \ln (c)

I understand everything except for the final term of the last equation. where did the \frac{1}{4} \ln (c) come from and shouldn't it just be c as the constant of integration?

Homework Equations

\frac{1}{4} \frac {du}{(2-u)} \ + \ \frac{1}{4} \frac{du}{(2+u)} \ = \frac{dx}{x}
\frac{-1}{4} \ln (2-u) \ + \ \frac{1}{4} \ln (2+u) \ = \ln (x) + \frac{1}{4} \ln (c)

The Attempt at a Solution

Integrating \frac{dx}{x} gives you \ln(x) + c.

 

[pasmith]

\frac14 \ln c is an arbitrary constant. Presumably writing it in that form rather than as D makes solving for u much more convenient.

If f: U \to \mathbb{R} is a surjection, then it is perfectly legitimate to take f(C) for some C \in U \subset \mathbb{R} as a constant of integration rather than D \in \mathbb{R}.

 

[dchau503]

So then any term as the constant would make sense? E.g. the book could've used \frac{1}{8} \ln(c) \text{or}\ 3\ln(c) as the last term in the integrated equation? I'm just a little confused how they got \frac{1}{4} \ln(c).

 

[vela]

Try using a plain old $c$ for the arbitrary constant and solve for $u$. What happens then if you replace $c$ with $\frac 14 \ln c$? You should, I hope, see why that form was chosen.

 

[dchau503]

I got that u = \frac{2cx^4-2}{1+cx^4}. I still don't see why the form was chosen.