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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u ^{2}}{8} - e^{u} u d u

= \int \frac{e ^{u} u ^{2}}{8} - e^{u} u 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

= \int \frac{e ^{u} u ^{2}}{8} d u - \int e^{u} u d u

\int \frac{e ^{u} u ^{2}}{8} d u = \frac{1}{8} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

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

= \frac{1}{8} (u^{2} e^{u} - 2 (e^{u} u - e^{u})) - (e^{u} u - e^{u})

Substitute back

u = 2 x

= \frac{1}{8} ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x})) - (e^{2 x} \cdot 2 x - e^{2 x})

Simplify

\frac{1}{8} ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x})) - (e^{2 x} \cdot 2 x - e^{2 x}) : \frac{1}{4} e^{2 x} (2 x^{2} - 2 x + 1) - 2 e^{2 x} x + e^{2 x}

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1 0 ^{n}}{n ^{n}}

4 x^{4}

y \cdot e^{y} d x + (1 + x \cdot e^{y}) d y = 0, y (5) = 0, y = y

f (t) = 8 t - 9 t^{2}

\frac{\partial}{\partial x} (e^{x y} + 2 x - 5 y)

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