It's getting too long!

$\displaystyle\int x^4e^x \ dx = e^x\left(Ax^4 - Bx^3 + Cx^2 - Dx + E\right) + \small \color{grey}{constant}$

Differentiate both sides w.r.t x:-

$x^4e^x=e^x(Ax^4+(4A-B)x^3+(C-3B)x^2+(2C-D)x+E-D)$

Comparing coefficients , we get:- $A=1,B=4A=4,C=3B=12,E=D=2C=24$ Hence, $A+B+C+D+E=\large\boxed{65}$

Use tabular integration by parts.

$x^4 \quad e^x$ $4x^3 \quad e^x$ $12x^2 \quad e^x$ $24x \quad e^x$ $24 \quad e^x$ $0 \quad e^x$

You multiply diagonally and alternate the signs.

This gives: $x^4 e^x - 4x^3 e^x + 12x^2 e^x - 24x e^x + 24 e^x + C$

Summing together the absolute value of all coefficients (which was what was asked for ) is $1 + 4 + 12 + 24 + 24 = \boxed{65}$

By integration by parts:

$\begin{aligned} I & = \int x^4e^x \ dx \\ & = x^4 e^x - \int 4x^3e^x \ dx \\ & = x^4 e^x - 4x^3e^x + \int 12x^2e^x \ dx \\ & = x^4 e^x - 4x^3e^x + 12x^2e^x - \int 24xe^x \ dx \\ & = x^4 e^x - 4x^3e^x + 12x^2e^x - 24xe^x + \int 24e^x \ dx \\ & = x^4 e^x - 4x^3e^x + 12x^2e^x - 24xe^x + 24e^x + \small \color{grey}{constant} \\ & = e^x \left(x^4 - 4x^3 + 12x^2 - 24x + 24 \right) + \small \color{grey}{constant} \end{aligned}$

$\implies A+B+C+D+E = 1+4+12+24+24 = \boxed{65}$

$\begin{aligned} \int x^4e^x \ dx & = x^4\int e^x \ dx - \int\left[\dfrac{d}{dx} x^4 × \int e^x \ dx\right] \ dx\\ \\ & = x^4e^x - \int\left[4x^3e^x\right] \ dx\\ \\ & = x^4e^x - \left[4x^3\int e^x \ dx - \int\left[\dfrac{d}{dx} 4x^3 × \int e^x \ dx\right] \ dx\right]\\ \\ & = x^4e^x - \left[4x^3e^x - \int\left[12x^2e^x\right]\ dx\right]\\ \\ & = x^4e^x - \left[4x^3e^x - \left[12x^2\int e^x \ dx - \int\left[\dfrac{d}{dx} 12x^2 × \int e^x \ dx\right] \ dx\right]\right]\\ \\ & = x^4e^x - \left[4x^3e^x - \left[12x^2e^x - \int\left[24xe^x\right] \ dx\right]\right]\\ \\ & = x^4e^x - \left[4x^3e^x - \left[12x^2e^x - \left[24xe^x - \int\left[24e^x\right] \ dx\right]\right]\right]\\ \\ & = x^4e^x - \left[4x^3e^x - \left[12x^2e^x - \left[24xe^x - \int\left[24e^x\right] \ dx\right]\right]\right]\\ \\ & = x^4e^x - \left[4x^3e^x - \left[12x^2e^x - \left[24xe^x - 24e^x\right]\right]\right]\\ \\ & = x^4e^x - 4x^3e^x + 12x^2e^x - 24xe^x + 24e^x + \small \color{grey}{constant}\\ \\ & = e^x\left(x^4 - 4x^3 + 12x^2 - 24x + 24\right) + \small \color{grey}{constant}\\ \\ \therefore a+b+c+d+e & = 1+4+12+24+24\\ & = \color{#69047E}{\boxed{65}} \end{aligned}$

This method works when you have a product of an exponential and a polynomial because integrating exponentials is typically easy as well as differentiating polynomials. As you can see, we differentiate the polynomial until it's a constant term, and then sum the products of the down-right diagonals (as you can see in the picture). You can even generalize this for the integral of x^n*e^x dx fairly easily, where n is in the positive integers.

By using Tabular.