Question

Partial Fraction Decomposition:\[\frac{(x+1)^2}{(x+2)^4}\]How can I do it?

Answer 1

\[\frac{A}{x+2}+\frac{B}{(x+2)^2}+\frac{C}{(x+2)^3}+\frac{D}{(x+2)^4}=\frac{(x+1)^2}{(x+2)^4}.\]Is this first step correct?

Answer 2

\[A(x+2)^3+B(x+2)^2+C(x+2)+D=(x+1)^2.\]

Answer 3

\[D=1.\]

Answer 4

Then I get stuck on how to find the rest of the variables.

Answer 5

when x=-1, \[A+B+C=-1\] and when x=3 \[-A+B-C=-2\] if you sum these equations you get B=-3/2

Answer 6

hero, whats your idea here?

Answer 7

when x=0 \[8A+4B+2C=0\] and \[C=3-4A\] if you put it to equation wher x=-1, you ge A=5/6

Answer 8

\[A+B+C=-1,\]\[-A+B-C=3,\]\[2B=2\implies B=1,\]... after doing some math...\[8A+2B=2\implies 8A+2=2\implies A=0,\]\[0+1+C=-1\implies C=-2\]... therefore...
\[\frac{1}{(x+2)^2}-\frac{2}{(x+2)^3}+\frac{1}{(x+2)^4}=\frac{(x+1)^2}{(x+2)^4}?\]

Answer 9

If you have access to Mathematica 8, select the problem expression with the mouse and click on the "Apart" button located in the Algebraic palette.\[\frac{1}{(x+2)^2}-\frac{2}{(x+2)^3}+\frac{1}{(x+2)^4} \]

Answer 10

So that's the answer. Thank you, robtobey, for the corroboration, and thank you, mathmagician, for the huge hint!

Answer 11

you are welcome, but still i mad a mistake...

Answer 12

i think its like this
(x+1)^2 (x+2)^4 =A1/(x+1) +A2/(x+1)^2 + A3/(x+2)+A4/(x+2)^2 +A5/(x+2)^3 +A6/(x+2)^4

Answer 13

1=[A1/(x+1) +A2/(x+1)^2 + A3/(x+2)+A4/(x+2)^2 +A5/(x+2)^3 +A6/(x+2)^4](x+1)^2 (x+2)^4

Answer 14

oh yes sorry

Answer 15

1=[A1(x+1) (x+2)^4 +A2 (x+2)^4 + A3(x+1)^2 (x+2)^3+A4(x+1)^2 (x+2)^2 +A5(x+1) (x+2)+A6(x+1)^2