Evaluate the sum: $\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+2)\ n!}$ 
How to evaluate the below sum?
$$\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+2)\ n!}$$


I was trying to find the integral of $x\ e^x$ without using Integral By Parts.
Here's what I got so far:
$$\begin{aligned}e^x &= \sum_{n=0}^\infty \dfrac{x^n}{n!}\\x\ e^x &= \sum_{n=0}^\infty \dfrac{x^{n+1}}{n!} \\\int x\ e^x\ dx& = \int\sum_{n=0}^\infty \dfrac{x^{n+1}}{n!} dx\\& = \sum_{n=0}^\infty \int\dfrac{x^{n+1}}{n!} dx\\& = \sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2)\ n!}  + C\end{aligned}$$
How to evaluate this series? I know the answer should be $e^x(x-1) + c$ which is the antiderivative of $x \ e^x$ and I've verified it from WolframAlpha, but not getting any idea how to evaluate the series.

I think partial fraction decomposition might work here because $\frac{1}{(n+2)n!} = \frac{1}{(n+1)!} - \frac{1}{(n+2)!}$ but not sure how to prove this and how to implement.


You have\begin{align}\frac1{(n+1)!}-\frac1{(n+2)!}&=\frac{n+2-1}{(n+2)!}\\&=\frac{\cancel{n+1}}{(n+2)\cancel{(n+1)}n!}\require{cancel}\\&=\frac1{(n+2)n!}\end{align}
and therefore you have\begin{align}\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)n!}&=\sum_{n=0}^\infty\left(\frac1{(n+1)!}-\frac1{(n+2)!}\right)x^{n+2}\\&=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+1)!}-\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}\\&=x\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)!}-\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}\\&=x(e^x-1)-(e^x-1-x)\\&=(x-1)e^x+1\end{align}
