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

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

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integral of (xe^{-2x})

\int (x e^{- 2 x}) d x

\frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + C

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

= \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - 2 x

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

Simplify

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

Add a constant to the solution

= \frac{1}{4} (- 2 e^{- 2 x} x - e^{- 2 x}) + C

y^{^{''}} + 16 y = 0, y (0) = 2, y^{^{'}} (0) = 1

\frac{\partial}{\partial x} (arcsin (5 x))

y^{^{''''}} + 4 y^{^{''}} = 0

x \to 1 lim (x ln (x) - x)

\frac{\partial}{\partial x} ((x + y) e^{y})

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