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Integrate 1/(3-e^x) using the substitution u=3-e^x the answer is what i got but with 3-e^x instead
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i got the answer \[1/3(\ln((-e^x)/(-3+e^x)))\]
i got ln(3-ex)+x
divide that by 3...sry
erm... when i differentiate that i got \[(-2e^x+3)/(3-e^x)\] which is not what is given by the question
divide that by 3 srry
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lol which one?
ur ans 1/3(ln(3-e^x)+x)
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