What is the integral of e^{(-3x^2-1)}6x ?

The integral of e^{(-3x^2-1)}6x is -e^{-3x^2-1}+C

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integral of e^{(-3x^2-1)}6x

\int e^{(- 3 x^{2} - 1)} 6 x d x

- e^{- 3 x^{2} - 1} + C

Solution steps

\int e^{- 3 x^{2} - 1} \cdot 6 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^{2} - 1} x d x

Apply u-substitution

: \int - \frac{e ^{u}}{6} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = - 3 x^{2} - 1

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

Simplify

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

= - e^{- 3 x^{2} - 1}

Add a constant to the solution

= - e^{- 3 x^{2} - 1} + C

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What is the integral of e^{(-3x^2-1)}6x ?
The integral of e^{(-3x^2-1)}6x is -e^{-3x^2-1}+C