What is the integral of xe^{(-4x)} ?

The integral of xe^{(-4x)} is 1/16 (-4e^{-4x}x-e^{-4x})+C

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integral of xe^{(-4x)}

\int x e^{(- 4 x)} d x

\frac{1}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

Solution steps

\int x e^{- 4 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{16} d u

= \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = - 4 x

= \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x})

Simplify

= \frac{1}{16} (- 4 e^{- 4 x} x - e^{- 4 x})

Add a constant to the solution

= \frac{1}{16} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

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What is the integral of xe^{(-4x)} ?
The integral of xe^{(-4x)} is 1/16 (-4e^{-4x}x-e^{-4x})+C