- #1
dchau503
- 13
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Homework Statement
The givenequation is this: [tex]
\frac{1}{4} \frac {du}{(2-u)} \ + \ \frac{1}{4} \frac{du}{(2+u)} \ = \frac{dx}{x}[/tex]
My book says that when integrated, the above equation becomes [tex]
\frac{-1}{4} \ln (2-u) \ + \ \frac{1}{4} \ln (2+u) \ = \ln (x) + \frac{1}{4} \ln (c) [/tex]
I understand everything except for the final term of the last equation. where did the [itex] \frac{1}{4} \ln (c) [/itex] come from and shouldn't it just be c as the constant of integration?
Homework Equations
[tex]\frac{1}{4} \frac {du}{(2-u)} \ + \ \frac{1}{4} \frac{du}{(2+u)} \ = \frac{dx}{x}[/tex]
[tex]\frac{-1}{4} \ln (2-u) \ + \ \frac{1}{4} \ln (2+u) \ = \ln (x) + \frac{1}{4} \ln (c) [/tex]
The Attempt at a Solution
Integrating [tex] \frac{dx}{x} [/tex] gives you [tex]\ln(x) + c[/tex].