What is the integral of (1-6x)e^{3x-9x^2} ?

The integral of (1-6x)e^{3x-9x^2} is 1/3 e^{3x(-3x+1)}+C

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integral of (1-6x)e^{3x-9x^2}

\int (1 - 6 x) e^{3 x - 9 x^{2}} d x

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

Solution steps

\int (1 - 6 x) e^{3 x - 9 x^{2}} d x

Apply u-substitution

: \int - \frac{e ^{\frac{- u ^{2} + 1}{4}} u}{6} d u

= \int - \frac{e ^{\frac{- u ^{2} + 1}{4}} u}{6} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int e^{\frac{- u ^{2} + 1}{4}} u d u

\frac{- u ^{2} + 1}{4} = - \frac{u ^{2}}{4} + \frac{1}{4}

= - \frac{1}{6} \cdot \int e^{- \frac{u ^{2}}{4} + \frac{1}{4}} u d u

Apply u-substitution

: \int - 2 e^{v} d v

= - \frac{1}{6} \cdot \int - 2 e^{v} d v

Take the constant out:

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

= - \frac{1}{6} (- 2 \cdot \int e^{v} d v)

Use the common integral:

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

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

Substitute back

= - \frac{1}{6} (- 2 e^{- \frac{( 1 - 6 x ) ^{2}}{4} + \frac{1}{4}})

Simplify

- \frac{1}{6} (- 2 e^{- \frac{( 1 - 6 x ) ^{2}}{4} + \frac{1}{4}}) : \frac{1}{3} e^{3 x (- 3 x + 1)}

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

Add a constant to the solution

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

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What is the integral of (1-6x)e^{3x-9x^2} ?
The integral of (1-6x)e^{3x-9x^2} is 1/3 e^{3x(-3x+1)}+C