What is the integral of xae^{-ax} ?

The integral of xae^{-ax} is -e^{-ax}x-(e^{-ax})/a+C

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integral of xae^{-ax}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

Substitute back

u = - a x

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

Simplify

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

= - e^{- a x} x - \frac{e ^{- a x}}{a}

Add a constant to the solution

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

What is the integral of xae^{-ax} ?
The integral of xae^{-ax} is -e^{-ax}x-(e^{-ax})/a+C