Find the indefinite integral:

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Find the indefinite integral:

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\int\limits_{?}^{?}\frac{ e^{y} }{ 2-e^{y} } dy\]
i know i'll be using substitution.
Let\[u=2-e^y\]then\[du=?\]

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e^y y dy?
\[u=2-e^y\implies du=-e^ydy\implies e^ydy=-du\]so what is the integral now?
\[\int\limits_{?}^{?} 1/du du\]
well minus?
how did you get two du's ?
\[\int\limits_{e^y}}^{2-e^{y}} e^{\]
\[\int\limits_{}^{} \frac{ e^{y} }{ 2-e{y} } e^{y}\]
where is the other e^y coming from?
thats my du.
well dy.
it should be negative.
is the answer ln|2-e^y|+c?
you dropped the negative sign, but aside from that the answer is correct

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