hey guys,
i follow you up to the point of intergrating the final line. How would I intergrate [x^-2.e^-x] dx
thx
3 answers
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 function 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/(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 :)
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 :)
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
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
0 answers