drwls & Damon  maths
posted by xX_Supaman_Xx on .
hey guys,
i follow you up to the point of intergrating the final line. How would I intergrate [x^2.e^x] dx
thx

My first suggestion was to try the method of "integration by parts", but that may not work. I found a recursion formula for the integral in my Table of Integrals, but it ends up requiring the integral of e^x/x , which is a new fucntion called the exponential integral, Ei(x). Perhaps I made a mistake somewhere applying the "integrating factor" method. Damon has made more progress and is better equipped to help you. I am giving up on it

Yes, use the recursion formula. You need it later to cancel the next term.
in general:
int e^ax dx/x^n = (1/(n1)) [e^ax/x^(n1) + a int e^ax/x^(n1)dx] for n integergreater than one
here n is two
so here I had:
y e^x/x=int( e^x dx/x)(1/x + 1)dx + C
ye^x/x =int e^x dx/x^2 + int e^x dx/x
+C
ye^x / x = (1/1)[e^x/x  int e^x dx/x ] + e^x dx/x +C
see that gets rid of the int e^x dx/x terms and you are left with
y e^x/x = e^x/x + C
y = 1 + x C / e^x
y = x C e^x  1 :) 
Typo:
ye^x / x = (1/1)[e^x/x  int e^x dx/x ] + INTEGRAL e^x dx/x +C
see that gets rid of the int e^x dx/x terms and you are left with