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...

[tex]\int \frac{1}{1-y} dy = - \int \frac{1}{u} du[/tex]

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

bmed90

gotcha

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bmed90	Why do you Tack on The Negative For this Integral 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 Recognitions: Gold Member	Because using u-substitution, we get this: u = 1 - y, du = - dy And so... [tex]\int \frac{1}{1-y} dy = - \int \frac{1}{u} du[/tex] Integrating the right-hand side, we get - ln\|u\| + C, or - ln\|1-y\| + C.