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

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

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

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

x + e^{- x} + C

Solution steps

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

Apply u-substitution

: \int - 1 + e^{u} d u

= \int - 1 + e^{u} d u

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 u + \int e^{u} d u

\int 1 d u = u

\int e^{u} d u = e^{u}

= - u + e^{u}

Substitute back

u = - x

= - (- x) + e^{- x}

Simplify

= x + e^{- x}

Add a constant to the solution

= x + e^{- x} + C

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