What is the integral of xe^{-0.4x} ?

The integral of xe^{-0.4x} is 6.25(-0.4e^{-0.4x}x-e^{-0.4x})+C

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integral of xe^{-0.4x}

\int x e^{- 0.4 x} d x

6.25 (- 0.4 e^{- 0.4 x} x - e^{- 0.4 x}) + C

Solution steps

\int x e^{- 0.4 x} d x

Apply u-substitution

: \int \frac{5 e ^{u} u}{0.8} d u

= \int \frac{5 e ^{u} u}{0.8} d u

Take the constant out:

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

= \frac{5}{0.8} \cdot \int e^{u} u d u

Simplify

= 6.25 \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 6.25 (e^{u} u - \int e^{u} d u)

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

= 6.25 (e^{u} u - e^{u})

Substitute back

u = - 0.4 x

= 6.25 (e^{- 0.4 x} (- 0.4 x) - e^{- 0.4 x})

Simplify

= 6.25 (- 0.4 e^{- 0.4 x} x - e^{- 0.4 x})

Add a constant to the solution

= 6.25 (- 0.4 e^{- 0.4 x} x - e^{- 0.4 x}) + C

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What is the integral of xe^{-0.4x} ?
The integral of xe^{-0.4x} is 6.25(-0.4e^{-0.4x}x-e^{-0.4x})+C