text: 1) Let be I(n)= integral of (x^n)(e^x)\nThen I(n)=(x^n)*e^x - n*I(n-1)\ngives a recursive formula.\n\n2) You can also write:\nI(n)=(e^x)*Sum(k=0 to n of\n(x^(n-k))*(n!/(n-k)!)*(-1)^k )\n\n3) As e^x=Sum(k=0 to infinity of (x^k)/k!)\nyou can write it also as an infinite sum:\nI(n)=(x^(n+1))*Sum(k=0 to infinity of\n(x^k)/(k!*(n+k+1))\n\nYou can easily prove it by simple derivation.\n(0!=1)\n\nI hope this can help