Notation:
The nth power of y is written y^n.
The nth derivative of y is written y@n.
The first derivative of y is written y @ 1 or y'
The second derivation of y is written y @ 2 or y''
etc
Note the following parallel between ^ and @
(x + y) ^ n (u v) @ n
( x + y) ^ 0 = x^0 y^0 = 1 (u v ) @ 0 =( u @ 0) ( v @ 0) = u v
(x + y ) ^1 = x^1 y^0 + y^1 x^0 (u v) @ 1 = u @ 1 v @ 0 + u @ 0 v @ 1
= x + y = u' v + u v'
(x + y) ^ 2 = x^2 y^0 + 2 x^1 y^1 + x^0 y^2
= x^2 + 2 x y + y^2
(u v )@ 2 = u @ 2 v @ 0 + 2 u @ 1 v @ 1 + u @ 0 v @ 2
= u'' v + 2 u' v' + u v''
(x + y) ^3 = x^3 y^0 + 3 x^2 y^1 + 3 x^1 y^2 + x^0 y^3
(u v) @ 3 = u@3 v@0 + 3 u@2 v@1 + 3 u@1 v@2 + u@0 v@3
= u''' v + 3 u'' v' + 3 u' v'' + u v'''
etc
Now, what does the formula for (x + y)^n give if we put n = -1?
(x + y) ^ (-1)
= x^(-1) y^0 - x^(-2) y^1 + x^(-3) y^2 - x^(-4) y^3 + ...
So the integral of the product of two functions can be given the formuala
( u v) @ (-1) = u@(-1) v - u@(-2) v' + u@(-3) v'' - u@(-4) v''' + ...
This makes it easy to integrate the product of an exponential function
with a polynomial. We can find any order integral of the exponential
function easily, and the succesive derivatives of the polynomial
eventually reach 0. Therefore we would identify the exponential function
with u and the polynomial function with v in the above formula.
Suppose we wanted to find the integral of x sin(x) ?
( u v) @ (-1) = u@(-1) v - u@(-2) v' + u@(-3) v'' - u@(-4) v''' + ...
Set u = sin(x) and v = x
u@(-1) = - cos(x) v' = 1
u@(-2) = - sin(x) v'' = 0
( x sin(x) ) @ (-1) = [ - cos(x)] x - [ - sin(x) ] [ 1 ] + 0
= - x cos(x) + sin(x)
And you can verify by taking derivative.
w = - x cos(x) + sin(x)
w' = - x [- sin(x) ] - cos(x) + cos(x)
= x sin(x)