That looks like an exponential.

We have $I = \int \dfrac{x^n}{1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}}dx$ Let $g(x,n)$ be defined as $g(x,n) = \ln \left( \sum_{k=0}^{n}\frac{x^k}{k!}\right)$ $\dfrac{\partial g(x,n)}{\partial x} = \dfrac{\displaystyle \sum_{k=0}^{n-1}\frac{x^k}{k!}}{\displaystyle \sum_{k=0}^{n}\frac{x^k}{k!}}$ $= 1 - \dfrac{1}{n!}\dfrac{x^n}{\displaystyle \sum_{k=0}^{n}\frac{x^k}{k!}}$ Or $\dfrac{x^n}{\displaystyle \sum_{k=0}^{n}\frac{x^k}{k!}} = n!(1-g'(x,n))$ Hence, $I = \int n!(1-g'(x,n))dx = n!(x - g(x,n)) + C$ So, $f(x,n) = n!(x- \ln\left(\sum_{k=0}^{n}\frac{x^k}{k!}\right))$ $\lim_{n \rightarrow \infty} \dfrac{f(x,n)}{n!} = x - \ln(e^x) = 0$