<s>[INST] Evaluating $\int{ \frac{x^n}{1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!}}}dx$ using Pascal inversion (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 ????
 [/INST] 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 

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

</s>