What is the integral of 12xe^{-3x} ?

The integral of 12xe^{-3x} is -4e^{-3x}x-4/3 e^{-3x}+C

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integral of 12xe^{-3x}

\int 12 x e^{- 3 x} d x

- 4 e^{- 3 x} x - \frac{4}{3} e^{- 3 x} + C

Solution steps

\int 12 x e^{- 3 x} d x

Take the constant out:

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

= 12 \cdot \int x e^{- 3 x} d x

Apply u-substitution

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

= 12 \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

= 12 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 12 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 12 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = - 3 x

= 12 \cdot \frac{1}{9} (e^{- 3 x} (- 3 x) - e^{- 3 x})

Simplify

12 \cdot \frac{1}{9} (e^{- 3 x} (- 3 x) - e^{- 3 x}) : - 4 e^{- 3 x} x - \frac{4}{3} e^{- 3 x}

= - 4 e^{- 3 x} x - \frac{4}{3} e^{- 3 x}

Add a constant to the solution

= - 4 e^{- 3 x} x - \frac{4}{3} e^{- 3 x} + C

\frac{d}{d x} (x co x (x))

f (x) = 2 + \frac{1}{x}, (4, 2.25)

\int_{- 2}^{2} (x^{3} - 2 x + 3) d x

\int (6 x^{2} y - 2 x) d y

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What is the integral of 12xe^{-3x} ?
The integral of 12xe^{-3x} is -4e^{-3x}x-4/3 e^{-3x}+C