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 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 :)
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
To integrate the function ∫[x^(-2) * e^(-x)] dx, you can use the technique of integration by parts. Integration by parts is based on the product rule for differentiation, but it is reversed for integration.
The integration by parts formula is ∫(u * dv) = u * v - ∫(v * du), where u and v are functions of x, and du and dv are their respective derivatives.
Let's assign the terms in your integral to u and dv:
u = x^(-2)
dv = e^(-x) dx
To find du, we differentiate u with respect to x:
du = d/dx (x^(-2)) dx
= -2 * x^(-3) dx
To find v, we integrate dv, which is e^(-x) dx:
v = ∫(e^(-x) dx)
= -e^(-x)
Now, we have all the elements we need to apply the integration by parts formula:
∫[x^(-2) * e^(-x)] dx = u * v - ∫(v * du)
= x^(-2) * (-e^(-x)) - ∫((-e^(-x)) * (-2 * x^(-3)) dx)
= -x^(-2) * e^(-x) + 2∫(x^(-3) * e^(-x)) dx
Now, we have a new integral to solve: ∫(x^(-3) * e^(-x)) dx
This integral can be evaluated using integration by parts again. You can follow the same procedure as before, assigning x^(-3) as u and e^(-x) as dv.
By using integration by parts twice, you will eventually reach a point where you have integrals that can be easily solved or simplified.
Keep in mind that integration by parts may require multiple iterations, depending on the complexity of the integral.