Question

Integration by Partial Fractions: Repeated Linear Factors

Answer 1

\[\int\limits \frac{ 1 }{ (x^2-1)^2 }\]

Answer 2

I've separated it into:

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

I set x = 1/-1 and found D and B = 1/4. How do I find the other two unknowns? Stuck at this one. Any help appreciated.

Answer 3

your denominators a quadratic, so Ax^3 + Bx^2 + Cx + D

Answer 4

\[\frac{1}{(x^2-1)^2}=\frac{Ax^3 + Bx^2 + Cx + D}{(x^2-1)^2}\]

let x=0
\[1=\frac{D}{1}~:~D=1\]
\[\frac{1}{(x^2-1)^2}=\frac{Ax^3 + Bx^2 + Cx + 1}{(x^2-1)^2}\]
use 3 different values and you can setup a system of 3 equaions now

Answer 5

I thought only if the quadratic denominator can't be factored then only can the above convention be used?

Answer 6

well it does have its drawbacks i spose

Answer 7

at the moment ABC are 0

Answer 8

Attached is wolfram's calculation. for the partial fraction part.

Answer 9

difference of squares .... totally didnt see that

Answer 10

Do we really have to expand this all out?
Dang that's not going to be fun...

Answer 11

Sorry, gotta go for abit. Will be back in about 20mins!

Answer 12

I'll post a pic of my working so far.

Answer 13

I guess another option, following what you've done so far,
umm..

plug in x=0, and x=2.. something like that.
And you end up with a system of 2 equations with A and C, ya?

Answer 14

If I solve the 2 equations I can't get the same answer that wolfram gave me, and consequently can't solve the integration part.

Answer 15

bbiab

Answer 16

1 +2x^2 +3x^4 +4x^6 + ...
--------------------
1 - 2x^2 +x^4 | 1
(1 - 2x^2 +x^4)
2x^2 -x^4
(2x^2 - 4x^4 +2x^6)
--------------------
3x^4 -2x^6
(3x^4 - 6x^6 +3x^8)
-------------------
4x^6 -3x^8



\[f'(x)=\sum_{n=0}(n+1)x^{2n}\]
\[\int f'(x)=\sum_{n=0}(n+1)\int x^{2n}\]
\[ f(x)=\sum_{n=0}\frac{n+1}{2n+1}x^{2n+1}\]

Answer 17

just playing around with it, the series agrees with the wolfs solution ...

Answer 18

Tbh I don't understand the above working lol... So from the wolfram solution A, B, C, and D all = 1/4?

Answer 19

They came up with C=-1/4 it looks like.

Answer 20

Oh yea. Missed that. I think I'll progress to a different question for now. Can't seem to solve it. Could be that my comparing of the coefficients was off, not sure.

Answer 21

I did x=0, solved for C in terms of A.
then x=2, and substituted for C.

It gave me the correct value for A. Seems to work out ok :O Just kind of long and tedious.

Answer 22

Ah! Ok got it. When I did x=0, I set C as + instead of -. Doh. Thanks! :)

Answer 23

Oh that (0-1)(0+1)^2 ?
Yah that got me at first also hehe

Answer 24

Yup that one! Haha...thanks for your help mate.