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

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

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integral of-2xe^{-2x}

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

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

Solution steps

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

Apply u-substitution

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

= - 2 \cdot \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

= - 2 \cdot \frac{1}{4} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= - 2 \cdot \frac{1}{4} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - 2 x

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

Simplify

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

= e^{- 2 x} x + \frac{1}{2} e^{- 2 x}

Add a constant to the solution

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

\frac{1}{2} t^{2} - 6 t + 23

\int (x^{3} ln (x)) d x

\int \frac{a ^{x}}{x} d x

\frac{d}{d θ} (ln (θ + 1))

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

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