How do you integrate #e^x/e^(2x+1)#?

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Answer 1 · 2016-08-03T21:44:21

#= - e^-(x+1) + C#

for starters

#e^x/e^(2x+1) = e^x *e^-(2x+1) = e^-(x+1)#

and as #d/dx e^(f(x)) = f'(x) e^(f(x))#

you are therefore looking at

#int \ e^-(x+1) \ dx #

#int \- d/dx ( e^-(x+1)) \ dx #

#= - e^-(x+1) + C#

of course, for #int \ e^-(x+1) \ dx #

you can introduce a sub like #u = -(x+1), du = - dx# and solve it as

#int \ - e^u \ du = - e^u + C = - e^-(x+1) + C#

if you prefer.