Posted by **xX_Supaman_Xx** on Saturday, December 29, 2007 at 9:45pm.

hey guys,

i follow you up to the point of intergrating the final line. How would I intergrate [x^-2.e^-x] dx

thx

- drwls & Damon - maths -
**drwls**, Sunday, December 30, 2007 at 10:20am
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

- drwls & Damon - maths -
**Damon**, Sunday, December 30, 2007 at 1:55pm
Yes, use the recursion formula. You need it later to cancel the next term.

in general:

int e^ax dx/x^n = (1/(n-1)) [-e^ax/x^(n-1) + a int e^ax/x^(n-1)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 :)

- drwls & Damon - maths -
**Damon**, Sunday, December 30, 2007 at 2:00pm
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

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