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