What is the integral of (ke^{-kx}) ?

The integral of (ke^{-kx}) is -e^{-kx}+C

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integral of (ke^{-kx})

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

- e^{- k x} + C

Solution steps

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

Take the constant out:

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

= k \cdot \int e^{- k x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{k} d u

= k \cdot \int - \frac{e ^{u}}{k} d u

Take the constant out:

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

= k (- \frac{1}{k} \cdot \int e^{u} d u)

Use the common integral:

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

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

Substitute back

u = - k x

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

Simplify

k (- \frac{1}{k} e^{- k x}) : - e^{- k x}

= - e^{- k x}

Add a constant to the solution

= - e^{- k x} + C

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