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)