General Formula

Quote:

Originally Posted by nick898

I'm looking for the general formula of the integral of (x^n*e^x)dx

Use integration by parts. Using the tabular method, with $u=x^n$ and $dv=e^x\,dx,$ we have

$\begin{tabular}{ccccc} Alternating&&$u$ and its&&$v$ and its\\ signs&&derivatives&&antiderivatives\\\hline +&$\rightarrow$&$x^n$&$_\searrow$&$e^x$\\ -&$\rightarrow$&$nx^{n-1}$&$_\searrow$&$e^x$\\ +&$\rightarrow$&$n(n-1)x^{n-2}$&$_\searrow$&$e^x$\\ -&$\rightarrow$&$n(n-1)(n-2)x^{n-3}$&$_\searrow$&$e^x$\\ $\vdots$&$\vdots$&$\vdots$&$\vdots$&$\vdots$\\ $+(-1)^n$&$\rightarrow$&$n!$&$_\searrow$&$e^x$\\ &&&&$e^x$ \end{tabular}$

So the formula should be

$\int x^ne^x\,dx=\sum_{i=0}^n\frac{(-1)^in!\,x^{n-i}e^x}{(n-i)!}$

$=n!\,e^x\sum_{i=0}^n\frac{(-1)^ix^{n-i}}{(n-i)!}$

$=x^ne^x-nx^{n-1}e^x+n(n-1)x^{n-2}e^x-\cdots\mp n!\,xe^x\pm n!\,e^x,$

where the last term is positive for even $n$ and negative for odd $n.$