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How do you integrate #int (5-e^x)/(e^(2x))dx#?

1 Answer

Dec 3, 2016

#-5/2e^(-2x)+e^-x+C#

Explanation:

rearrange as folows.

#int((5-e^x)/e^(2x))dx#

#=int(5/e^(2x)-e^x/e^(2x))dx#

#int(5e^(-2x)-e^(-x))dx#

#=-5/2e^(-2x)+e^-x+C#

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