Partial fraction

Let me guess -- you want us to do totally your homework for you without showing anything you have done to try and solve ther problem on your own? Is that about right?

No, its not that i hv done it but i'm lazy to type my working coz 2 long..My working is wrong so i jz wan 2 knw anybody knw alternate method to solve it..Jz doing some research..If got new method i learn more..If nt nevermind..Anyway my teacher wil explain it..Dont u think by doing this u might find easier ways to solve this questions than ur tcer method..I'm juz doing comparing..No offence..Sorry if i hv say o done anything wrng

In general, a fraction with n linear factors of x+1 in the denominator can be decomposed as follows:

A1/(x + 1) + A2/(x+1)^2 +...+ An/(x+1)^n

How come you didn't say "Please"?

Ok please anybody o miller can you all have a look what i have done..(x^2+23)/((x+1)^3(x-2))=a/(x+1)^3+ b/(x+1)^2 + c/(x+1) + d/(x-2)

=a(x-2) +b(x+1)(x-2)+ c(x+1)^2(x-2)+ d(x+1)^3

=(c+d)x^3+ (b+3d)x + (a-b-3c+3d)x^2 + (-2a-2b-2c+d)

comparing coefficient:

c+d=0 ->eqn 1, a-b-3c+3d=0 ->eqn 2, -2a-2b-2c+d=23 ->eqn 3, b+3d=1 ->eqn 4

put eqn 2=eqn 3: 3a+b-c+2d-23=0 ->eqn 5

put eqn 1=eqn 4: c+d= b+3d-1

-2d+ c- b+ 1=0 ->eqn 6

eqn 5 + eqn 6:

3a-22=0

3a=22

a=(22)/3

put a into eqn 5, b- c+ 2d-1=0

put eqn 1 into eqn 5 where c= -d, b-(-d)+2d-1=0

b+3d=1

please anybody know how to continue the rest & where i got wrong because the real answer is 1/(x-2) - 1/(x+1) -2/(x+1)^2 -8/(x+1)^3