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

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

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integral of e^{-2x}x

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

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

Solution steps

\int e^{- 2 x} 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

\frac{\partial}{\partial x} (x^{4} y^{8} + 4 x^{5} y)

ln (\frac{x}{x ^{2} + 2})

\frac{d}{d x} (5 lo g_{2} (x))

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 0, y (\frac{π}{2}) = 1

\frac{8}{( s - 1 ) ^{3}}

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