Intergration, Cover up rule

[Crash]

Im having some difficulty in grasping how to do the cover up rule with 3 factors as the denominator,

Like (x-2)/(x+1)(x+2)(x+4)

For the x+1 factor would i put -2 in for the (x+4), -4 in for the (x-2) factor and that would solve for the numerator, how do i do it, can someone give me a good explantaion on how to do it

Cheers

[lurflurf]

O.K.

First realize that as niffty as this method is it will not work if without change if their are repeated of nonlinear factors.

 

The idea is we know from a theorem in algebra that there exit numbers a,b,c so that

(x-2)/[(x+1)(x+2)(x+4)]=a/(x+1)+b/(x+2)+c/(x+4)

the cover up method works like this, take a number that makes the denominator zero such as -1 in this example. replace all the x's except the one in (x+1) with -1

(x'-2)/[(x+1)(x'+2)(x'+4)] x'->-1

(-1-2)/[(x+1)(-1+2)(-1+4)]=(-3)/[(x+1)(1)(3)]=-1/(x+1)

do the same thing with -2 and -4

(-2-2)/[(-2+1)(x+2)(-2+4)]=(-4)/[(-1)(x+2)(2)]=2/(x+2)

(-4-2)/[(-4+1)(-4+2)(x+4)]=(-6)/[(-3)(-2)(x+4)]=-1/(x+4)

the sum of these three equals the original function

(x-2)/[(x+1)(x+2)(x+4)]=-1/(x+1)+2/(x+2)-1/(x+4)

This new form is much easier to integrate.