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

The integral of (x^2-3x)e^{2x} is 1/2 e^{2x}(x^2-4x+2)+C

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

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

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

Solution steps

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

Apply u-substitution

= \int \frac{1}{2} e^{u} (\frac{u ^{2}}{4} - \frac{3 u}{2}) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int e^{u} (\frac{u ^{2}}{4} - \frac{3 u}{2}) d u

Expand

e^{u} (\frac{u ^{2}}{4} - \frac{3 u}{2}) : \frac{e ^{u} u ^{2}}{4} - \frac{3 e ^{u} u}{2}

= \frac{1}{2} \cdot \int \frac{e ^{u} u ^{2}}{4} - \frac{3 e ^{u} u}{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{e ^{u} u ^{2}}{4} d u - \int \frac{3 e ^{u} u}{2} d u)

\int \frac{e ^{u} u ^{2}}{4} d u = \frac{1}{4} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

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

= \frac{1}{2} (\frac{1}{4} (u^{2} e^{u} - 2 (e^{u} u - e^{u})) - \frac{3}{2} (e^{u} u - e^{u}))

Substitute back

u = 2 x

= \frac{1}{2} (\frac{1}{4} ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x})) - \frac{3}{2} (e^{2 x} \cdot 2 x - e^{2 x}))

Simplify

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

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

Add a constant to the solution

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

What is the integral of (x^2-3x)e^{2x} ?
The integral of (x^2-3x)e^{2x} is 1/2 e^{2x}(x^2-4x+2)+C