Evaluating $\int{ \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}}dx$ using Pascal inversion

You may observe that $$ \left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'=1 + x + \frac{x^2}{2} + \cdots + \frac{x^{n-1}}{(n-1)!} $$ giving $$ \left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)-\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'=\frac{x^n}{n!} $$ and $$ \begin{align} \int \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}dx&=n!\int\frac{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)-\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}\:dx\\\\ &=n!\int dx-n!\int\frac{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)'}{\left(1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right)}\:dx. \end{align} $$ Thus

$$ \int \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}dx=n!\:x-n!\ln \left| 1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}\right|+C. $$