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

sin (x) sec (y) d x = d y

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

\frac{d}{d x} (\frac{1 + x}{x})

\frac{d}{d x} ((a - \frac{b}{x})^{2})

\frac{2.4}{5 + 1 x}

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