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

The integral of x^4e^{2x} is 1/4 e^{2x}(2x^4-4x^3+6x^2-6x+3)+C

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integral of x^4e^{2x}

\int x^{4} e^{2 x} d x

\frac{1}{4} e^{2 x} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C

Solution steps

\int x^{4} e^{2 x} d x

Apply u-substitution

: \int \frac{e ^{u} u ^{4}}{32} d u

= \int \frac{e ^{u} u ^{4}}{32} d u

Take the constant out:

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

= \frac{1}{32} \cdot \int e^{u} u^{4} d u

Apply Integration By Parts

: u^{4} e^{u} - \int 4 u^{3} e^{u} d u

= \frac{1}{32} (u^{4} e^{u} - \int 4 u^{3} e^{u} d u)

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

= \frac{1}{32} (u^{4} e^{u} - 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))))

Substitute back

u = 2 x

= \frac{1}{32} ((2 x)^{4} e^{2 x} - 4 ((2 x)^{3} e^{2 x} - 3 ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x}))))

Simplify

\frac{1}{32} ((2 x)^{4} e^{2 x} - 4 ((2 x)^{3} e^{2 x} - 3 ((2 x)^{2} e^{2 x} - 2 (e^{2 x} \cdot 2 x - e^{2 x})))) : \frac{1}{4} e^{2 x} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3)

= \frac{1}{4} e^{2 x} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3)

Add a constant to the solution

= \frac{1}{4} e^{2 x} (2 x^{4} - 4 x^{3} + 6 x^{2} - 6 x + 3) + C

\int \frac{2 x + 1}{( 3 x - 1 ) ( x ^{2} + 2 x + 2 )} d x

\frac{\partial}{\partial x} (4 - 2 \cdot (3 x - y)^{2})

y = - (- 6 x + 6)^{\frac{1}{2}}, x = - 5

\frac{d}{d v} (a e^{v} + \frac{b}{v} + \frac{c}{v ^{2}})

x y^{^{'}} - 2 y + y^{2} = 0

What is the integral of x^4e^{2x} ?
The integral of x^4e^{2x} is 1/4 e^{2x}(2x^4-4x^3+6x^2-6x+3)+C