What is the integral of e^{-3x}(-3x) ?

The integral of e^{-3x}(-3x) is e^{-3x}x+1/3 e^{-3x}+C

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integral of e^{-3x}(-3x)

\int e^{- 3 x} (- 3 x) d x

e^{- 3 x} x + \frac{1}{3} e^{- 3 x} + C

Solution steps

\int e^{- 3 x} (- 3 x) d x

Simplify

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

Take the constant out:

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

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

Apply u-substitution

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

= - 3 \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

= - 3 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= - 3 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = - 3 x

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

Simplify

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

= e^{- 3 x} x + \frac{1}{3} e^{- 3 x}

Add a constant to the solution

= e^{- 3 x} x + \frac{1}{3} e^{- 3 x} + C

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What is the integral of e^{-3x}(-3x) ?
The integral of e^{-3x}(-3x) is e^{-3x}x+1/3 e^{-3x}+C