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

[dchau503]

Homework Statement

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

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.