Constant of integration positive or negative?

I was solving some equation related to mechanics ... anyways I reached this step:

$\int dx=-10\int 1+\frac{20}{v-20}dv$

It's correct according to the answer key of the book. So the next step is:

$x=-10(v+20\ln|v-20|)+C$

But the answers says

$x+C=-10(v+20\ln|v-20|)$

...doesn't matter, right?

Since

$x+D=-10(v+20\ln|v-20|)+E$ and $D-E$ is an unknown constant, and so is $E-D$.

Anyways, then it's given that when $x=0$, $v=0$ so we substitute those in to get the value of the constant.

So if you try substituting it you'd get $200\ln20$ but if you do it in my form of the equation you get an opposite sign to that of the answers'!

$x=-10(v+20\ln|v-20|)+C$

$0=-10(20\ln|20|)+C$

$C=10(20\ln|20|)$

$x=-10(v+20\ln|v-20|)+200\ln20$

or

$x+C=-10(v+20\ln|v-20|)$

$0+C=-10(20\ln|20|)$

$C=-10(20\ln|20|)$

$x=-10(v+20\ln|v-20|)-200\ln20$

I'm confused.

...Here's the question if it would be of any benefit

A particle starts from rest and moving under the action of the force $F=(2-0.1v)$N. Given that the mass of the particle is 1 Kg. Find the displacement when $V=10$ m/s.

1 Answer

In the first case you find C=-200ln(20), so when you take it to the right side of the equation it must change sign. So it is the same in both cases.