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

f (x) = \frac{- 3}{x ^{4}}

\int_{4}^{69} \frac{1}{x ^{3} - 64} d x

x \to 0 lim (x^{2} + 4)

f (x) = e^{x} (3 x^{2} - 6 x + 6)

y^{^{'}} = (t + y)^{2} - 1, y (1) = 5

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