What is the integral of (x-2)e^{x^2-4x} ?

The integral of (x-2)e^{x^2-4x} is 1/2 e^{x^2-4x}+C

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integral of (x-2)e^{x^2-4x}

\int (x - 2) e^{x^{2} - 4 x} d x

\frac{1}{2} e^{x^{2} - 4 x} + C

Solution steps

\int (x - 2) e^{x^{2} - 4 x} d x

Apply u-substitution

: \int e^{u^{2} - 4} u d u

= \int e^{u^{2} - 4} u d u

Apply u-substitution

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

= \int \frac{e ^{v}}{2} d v

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} e^{v}

Substitute back

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

Simplify

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

= \frac{1}{2} e^{x^{2} - 4 x}

Add a constant to the solution

= \frac{1}{2} e^{x^{2} - 4 x} + C

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What is the integral of (x-2)e^{x^2-4x} ?
The integral of (x-2)e^{x^2-4x} is 1/2 e^{x^2-4x}+C