What is the integral of 2xe^{-4x} ?

The integral of 2xe^{-4x} is 1/8 (-4e^{-4x}x-e^{-4x})+C

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integral of 2xe^{-4x}

\int 2 x e^{- 4 x} d x

\frac{1}{8} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

Solution steps

\int 2 x e^{- 4 x} d x

Take the constant out:

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

= 2 \cdot \int x e^{- 4 x} d x

Apply u-substitution

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

= 2 \cdot \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

= 2 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= 2 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = - 4 x

= 2 \cdot \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x})

Simplify

2 \cdot \frac{1}{16} (e^{- 4 x} (- 4 x) - e^{- 4 x}) : \frac{1}{8} (- 4 e^{- 4 x} x - e^{- 4 x})

= \frac{1}{8} (- 4 e^{- 4 x} x - e^{- 4 x})

Add a constant to the solution

= \frac{1}{8} (- 4 e^{- 4 x} x - e^{- 4 x}) + C

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What is the integral of 2xe^{-4x} ?
The integral of 2xe^{-4x} is 1/8 (-4e^{-4x}x-e^{-4x})+C