What is the integral of-kxe^{-kx} ?

The integral of-kxe^{-kx} is e^{-kx}x+(e^{-kx})/k+C

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integral of-kxe^{-kx}

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

e^{- k x} x + \frac{e ^{- k x}}{k} + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{k ^{2}} d u

= - k \cdot \int \frac{e ^{u} u}{k ^{2}} d u

Take the constant out:

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

= - k \frac{1}{k ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= - k \frac{1}{k ^{2}} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - k x

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

Simplify

- k \frac{1}{k ^{2}} (e^{- k x} (- k x) - e^{- k x}) : e^{- k x} x + \frac{e ^{- k x}}{k}

= e^{- k x} x + \frac{e ^{- k x}}{k}

Add a constant to the solution

= e^{- k x} x + \frac{e ^{- k x}}{k} + C

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What is the integral of-kxe^{-kx} ?
The integral of-kxe^{-kx} is e^{-kx}x+(e^{-kx})/k+C