What is the integral of g/k (1-e^{-kt}) ?

The integral of g/k (1-e^{-kt}) is g/k (t+e^{-kt} 1/k)+C

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integral of g/k (1-e^{-kt})

\int \frac{g}{k} (1 - e^{- k t}) d t

\frac{g}{k} (t + e^{- k t} \frac{1}{k}) + C

Solution steps

\int \frac{g}{k} (1 - e^{- k t}) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{g}{k} \cdot \int 1 - e^{- k t} d t

Apply the Sum Rule:

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

= \frac{g}{k} (\int 1 d t - \int e^{- k t} d t)

\int 1 d t = t

\int e^{- k t} d t = - \frac{1}{k} e^{- k t}

= \frac{g}{k} (t - (- \frac{1}{k} e^{- k t}))

Simplify

= \frac{g}{k} (t + e^{- k t} \frac{1}{k})

Add a constant to the solution

= \frac{g}{k} (t + e^{- k t} \frac{1}{k}) + C

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What is the integral of g/k (1-e^{-kt}) ?
The integral of g/k (1-e^{-kt}) is g/k (t+e^{-kt} 1/k)+C