What is the integral of+6e^{-3x} ?

The integral of+6e^{-3x} is -2e^{-3x}+C

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integral of+6e^{-3x}

\int + 6 e^{- 3 x} d x

- 2 e^{- 3 x} + C

Solution steps

\int 6 e^{- 3 x} d x

Take the constant out:

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

= 6 \cdot \int e^{- 3 x} d x

Apply u-substitution

: \int - \frac{1}{3} e^{u} d u

= 6 \cdot \int - \frac{1}{3} e^{u} d u

Take the constant out:

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

= 6 (- \frac{1}{3} \cdot \int e^{u} d u)

Use the common integral:

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

= 6 (- \frac{1}{3} e^{u})

Substitute back

u = - 3 x

= 6 (- \frac{1}{3} e^{- 3 x})

Simplify

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

= - 2 e^{- 3 x}

Add a constant to the solution

= - 2 e^{- 3 x} + C

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