(adsbygoogle = window.adsbygoogle || []).push({}); Does "+ C" melt together ALL constants?

I had this integral [tex]\int \frac{dy}{-2y + 6}[/tex].

I realize you can let [itex]u = -2y + 6[/itex] so that [itex]du = -2dy[/itex], and adjusting: [itex]-0.5 du = dy[/itex].

The integral becomes

[tex]-0.5 \int \frac{du}{u}[/tex]

[tex]= -0.5 \ln{|-2y + 6|} + C[/tex]

However, there is another way. Factor out the half before the integration:

[tex]\int \frac{dy}{-2y + 6}[/tex]

[tex]= \int \frac{dy}{-2(y - 3)}[/tex]

[tex]= -0.5 \int \frac{dy}{(y - 3)}[/tex]

[tex]= -0.5 \ln{|y - 3|} + C[/tex]

BUT!

[tex]-0.5\ln{|-2y + 6|} + C \neq -0.5\ln{|y - 3|} + C [/tex] !

INSTEAD,

[tex]-0.5\ln{|-2y + 6|} + C[/tex] equals [tex]-0.5\ln{|y - 3|} - 0.5\ln{2} + C[/tex] !

Can anybody explain this discrepancy? Does [itex]-0.5\ln{2} + C[/itex] turn simply into just [itex] + C[/itex]? I understand that there can be an infinite number of antiderivatives for a function. But this is just odd. Both integrations are correctly done, no? If, on a test, I gave one answer, where the other one was preferred, what would happen?

And yeah, both derivatives go back to being the same thing. Strange!

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# Does + C melt together ALL constants?

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