Why do you Tack on The Negative For this Integral

[bmed90]

If you integrate

1/(1-y)dy

why do you end up with a negative in front of your answer

-ln|1-y|+c

 

[Char. Limit]

Because using u-substitution, we get this:

u = 1 - y, du = - dy

And so...

$$\int \frac{1}{1-y} dy = - \int \frac{1}{u} du$$

Integrating the right-hand side, we get - ln|u| + C, or - ln|1-y| + C.

 

[bmed90]

gotcha