How do you integrate $(e^{2 x - 1}) - 1$ $(e^(2x-1))-1$ ?

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Answer 1 · 2016-07-20T14:45:00

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

$int \ (e^(2x-1))-1 \ dx$

$= int \ d/dx(1/2e^(2x-1))-1 \ dx$

$= 1/2e^(2x-1)- x + C$

How do you integrate (e^(2x-1))-1(e2x−1)−1 (e^(2x-1))-1?