How do you integrate $\frac{e^{x}}{e^{2 x + 1}}$ $e^x/e^(2x+1)$ ?

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Answer 1 · 2016-08-03T21:44:21

$= - e^{- (x + 1)} + C$ $= - e^-(x+1) + C$

for starters

$\frac{e^{x}}{e^{2 x + 1}} = e^{x} \cdot e^{- (2 x + 1)} = e^{- (x + 1)}$ $e^x/e^(2x+1) = e^x *e^-(2x+1) = e^-(x+1)$

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.

How do you integrate e^x/e^(2x+1)exe2x+1e^x/e^(2x+1)?