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)