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

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.