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

Take the constant out:

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

= - \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

\int (e) d x

x \to 0 lim (\frac{sin ( x )}{x} + 3)

f (x) = arccos (e^{x})

\frac{\partial}{\partial y} (x - y^{2})

x \to 0 lim (\frac{6 x ^{2}}{x ^{2}})

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