What is the integral of (1-e^{-2x}) ?

The integral of (1-e^{-2x}) is x+1/2 e^{-2x}+C

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integral of (1-e^{-2x})

\int (1 - e^{- 2 x}) d x

x + \frac{1}{2} e^{- 2 x} + C

Solution steps

\int 1 - e^{- 2 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x - \int e^{- 2 x} d x

\int 1 d x = x

\int e^{- 2 x} d x = - \frac{1}{2} e^{- 2 x}

= x - (- \frac{1}{2} e^{- 2 x})

Simplify

= x + \frac{1}{2} e^{- 2 x}

Add a constant to the solution

= x + \frac{1}{2} e^{- 2 x} + C

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What is the integral of (1-e^{-2x}) ?
The integral of (1-e^{-2x}) is x+1/2 e^{-2x}+C