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

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

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

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

\frac{1}{2} e^{3 x (x - 2)} + C

Solution steps

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

Apply u-substitution

= \int \frac{e ^{\frac{u ^{2} - 9}{3}} u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int e^{\frac{u ^{2} - 9}{3}} u d u

\frac{u ^{2} - 9}{3} = \frac{1}{3} (u^{2} - 9)

= \frac{1}{3} \cdot \int e^{\frac{1}{3} (u^{2} - 9)} u d u

Apply u-substitution

= \frac{1}{3} \cdot \int \frac{e ^{\frac{v}{3}}}{2} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int e^{\frac{v}{3}} d v

Apply u-substitution

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int e^{w} \cdot 3 d w

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot 3 \cdot \int e^{w} d w

Use the common integral:

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot 3 e^{w}

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} \cdot 3 e^{\frac{( 3 x - 3 ) ^{2} - 9}{3}}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot 3 e^{\frac{( 3 x - 3 ) ^{2} - 9}{3}} : \frac{1}{2} e^{3 x (x - 2)}

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

Add a constant to the solution

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

d er i v a t i v e y = 1 7^{x}

\frac{\partial}{\partial x} (sin (π (6 x - 5 y)))

d er i v a t i v e f (x) = \frac{3}{( x ^{2} - 5 x ) ^{2}}

d er i v a t i v e y = 1 cos (2) (x)

t a y l or e^{t^{2}}

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