A "trick"y Integral

Put $1+x+\frac {x^2}{2!}...\frac {x^4}{4!}=t$

Hence, $dt= (1+x+\frac {x^2}{2!} +\frac {x^3}{3!})dx$

Thus, $tdx-dt=\frac {x^4}{4!}dx$

Putting that in the integral,

$\Rightarrow \int \frac {24(tdx-dt)}{t}$

$=24\int {dx}+ 24\int \frac {dt}{t}$

Put $t$ in terms of $x$ and integrate, and we have the answer.

Original integral: $\int{\frac{x^4 dx}{1+x+\frac{x^2}{2!}+ . . . + \frac{x^4}{4!}}}$ .
Factoring out a 24 gives: $24\int{\frac{x^4 dx}{x^4+4x^3+12x^2+24x+24}}$ .
Performing long division gives: $\int{1-\frac{4x^3+12x^2+24x+24}{x^4+4x^3+12x^2+24x+24} dx}$ .
Separate the integrals: $\int{1 dx} - \int{\frac{4x^3+12x^2+24x+24}{x^4+4x^3+12x^2+24x+24} dx}$ .
Use U-Substitution where u is the denominator, and du is the numerator: $\int{1 dx} - \int{\frac{du}{u}}$ .
Integrate: $x - ln|u| + C$ .
Plug in u and bring back the original factor of 24: $24x-24ln(x^4+4x^3+12x^2+24x+24) + C$ .

$\begin{aligned} I & = \int \frac {x^4}{1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\frac {x^4}{4!}} \ dx & \small \color{#3D99F6} \text{Multiply up and down by }4! \\ & = \int \frac {24x^4}{24+24x+12x^2+4x^3+x^4} \ dx \\ & = \int \left(24 - \frac {24(24+24x+12x^2+4x^3)}{24+24x+12x^2+4x^3+x^4} \right) dx \\ & = \boxed{24x - 24 \ln (x^4+4x^3+12x^2+24x+24) + C} \end{aligned}$