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{d}{d x} (\frac{x ^{3} + 5}{x ^{2} + 2 x + 3})

f (x) = x^{3}

\int lo g_{6} (x) d x

\frac{\partial}{\partial x} (4 x y e^{- x^{2} - y^{2}})

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

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