What is the integral of 2xe^{-3x} ?

The integral of 2xe^{-3x} is 2/9 (-3e^{-3x}x-e^{-3x})+C

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integral of 2xe^{-3x}

\int 2 x e^{- 3 x} d x

\frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x}) + C

Solution steps

\int 2 x e^{- 3 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^{- 3 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{9} d u

= 2 \cdot \int \frac{e ^{u} u}{9} d u

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 2 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 2 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = - 3 x

= 2 \cdot \frac{1}{9} (e^{- 3 x} (- 3 x) - e^{- 3 x})

Simplify

2 \cdot \frac{1}{9} (e^{- 3 x} (- 3 x) - e^{- 3 x}) : \frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x})

= \frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x})

Add a constant to the solution

= \frac{2}{9} (- 3 e^{- 3 x} x - e^{- 3 x}) + C

\frac{d y}{d x} (y) = 7 - 9 tan (x)

f (x) = 3 x^{2} - 2 x

\frac{d y}{d x}, x y = 1

f (x) = x + \frac{1}{x}

x \to a + lim (\frac{cos ( x ) ln ( x - a )}{ln ( e ^{x} - e ^{a} )})

What is the integral of 2xe^{-3x} ?
The integral of 2xe^{-3x} is 2/9 (-3e^{-3x}x-e^{-3x})+C