What is the integral of xe^{-9x} ?

The integral of xe^{-9x} is 1/81 (-9e^{-9x}x-e^{-9x})+C

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integral of xe^{-9x}

\int x e^{- 9 x} d x

\frac{1}{81} (- 9 e^{- 9 x} x - e^{- 9 x}) + C

Solution steps

\int x e^{- 9 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{81} d u

= \int \frac{e ^{u} u}{81} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{81} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{81} (e^{u} u - e^{u})

Substitute back

u = - 9 x

= \frac{1}{81} (e^{- 9 x} (- 9 x) - e^{- 9 x})

Simplify

= \frac{1}{81} (- 9 e^{- 9 x} x - e^{- 9 x})

Add a constant to the solution

= \frac{1}{81} (- 9 e^{- 9 x} x - e^{- 9 x}) + C

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What is the integral of xe^{-9x} ?
The integral of xe^{-9x} is 1/81 (-9e^{-9x}x-e^{-9x})+C