text: what is the solution to the integral of....? what is the antiderivative or integral of:  (x^n)(e^x)dx where n = any positive integer and e is just natural e, 2.71.... i need a basic formula for any example.  for example (x^3,563)(e^x).  i know n can be infinitly large.  i'm seeing a pattern in the equation with n, so far i have as the answer:  =(x^n)(e^x)-[n(x^(n-1))(e^x)]+ [n(n-1)(x^(n-2))(e^x)] - [n(n-1)(n-2)(x^(n-3))(e^x)] ....    i've notice each time i subtract another number from n, i see this pattern n(n-1)(n-2)(n-3)(n-4)....(n-999)....(n-9999)...technically this equation goes on infinitly, but there's got to be some way to represent it? 1) let be i(n)= integral of (x^n)(e^x) then i(n)=(x^n)*e^x - n*i(n-1) gives a recursive formula.  2) you can also write: i(n)=(e^x)*sum(k=0 to n of (x^(n-k))*(n!/(n-k)!)*(-1)^k )  3) as e^x=sum(k=0 to infinity of (x^k)/k!) you can write it also as an infinite sum: i(n)=(x^(n+1))*sum(k=0 to infinity of (x^k)/(k!*(n+k+1))  you can easily prove it by simple derivation. (0!=1)  i hope this can help