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

Question

(Note: I apreciate very much who marked this as a duplicate but I would like an answer for why my proof is wrong)

This is my solution, I have no clue why it failed. Let's start:

define $$I_n(m) = \int_{0}^{x} \frac{t^m}{1 + t + t^2/2 + ... + t^n/n!}\ dt$$

so it should be true that

$$\sum_{m=0}^{n}\frac{I_n(m)}{m!} = x$$

Then I use Pascal inversion:

$$\sum_{m=0}^n \frac{n! I_n(m)}{m!} = n!x$$

$$\sum_{m=0}^n {n\choose m} B_n(m) = n!x$$

where $B_n(m) = (n-m)!I_n(m)$

by Pascal's formula:

$$I_n(n) = (-1)^nxn! \sum_{m=0}^{n} \frac{(-1)^m}{(n-m)!}$$

what did I do wrong ????

Answer 1

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

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