Why do you Tack on The Negative For this Integral

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SUMMARY

The integral of the function 1/(1-y) with respect to y results in a negative logarithmic expression, specifically -ln|1-y| + C. This outcome arises from the u-substitution method where u is defined as 1 - y, leading to du = -dy. Consequently, the integral transforms to -∫(1/u) du, which integrates to -ln|u| + C, ultimately yielding the negative sign in the final answer.

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bmed90
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If you integrate

1/(1-y)dy

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

-ln|1-y|+c
 
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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.
 
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