Question

another one

Answer 1

\[\int\limits_{ }^{ } \frac{1}{(x^2-1)^2}~dx\]

Answer 2

\[\int\limits_{ }^{ } \frac{1}{(x-1)^2(x+1)^2}~dx\]

Answer 3

\[\frac{A}{(x-1)^2}+\frac{B}{(x+1)^2}=\frac{1}{(x-1)^2(x+1)^2}\]

Answer 4

PARTIAL FRACTION

Answer 5

yes,\[A(x+1)^2+B(x-1)^2=1\]

Answer 6

so whats the problem??

Answer 7

when x=1..... A=1/4
when x=-1....B=1/4

Answer 8

I am working the practice problems and If I get mistkaes there are those that can check me

Answer 9

Your second step is flawed I think. Check your factoring

Answer 10

I used difference of squares, and then I know that
\[[~(c_1)(c_2)~]^2=(c_1)^2(c_2)^2\]

Answer 11

or where is the mistake?

Answer 12

MISTAKE LOL

Answer 13

wait lemme post

Answer 14

Nevermind, I misread the question

Answer 15

there was nothing to read.

Answer 16

just an integral, lol:P

Answer 17

\[A/(1+x) +B/(1+x)^2 \]

Answer 18

similarly for x-1

Answer 19

\[\int\limits_{ }^{ } \frac{1}{(x-1)^2(x+1)^2}~dx=\frac{1}{4}\int\limits_{ }^{ } \left(\begin{matrix} \frac{1}{(x+1)^2}+\frac{1}{(x-1)^2}\\ \end{matrix}\right)~dx\]so far....

Answer 20

GUYS U GOT LUCKY

Answer 21

then I can set u=x+1, and s=x-1 to be correct show-work wise, but there is no need actually,
because dx would be equal to du

I get....just using power rule,
\[-\frac{1}{4(x-1)}-\frac{1}{4(x+1)}+C\]

Answer 22

MY METHOD IS ACTUAL

Answer 23

http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Procedure

Answer 24

check the last point

Answer 25

my answer is wrong or correct?

Answer 26

again U got lucky I think

Answer 27

why

Answer 28

I didnt check urs

Answer 29

U m8 be rt

Answer 30

oh, I am not forcing you to

Answer 31

ur method is wrong I guess

Answer 32

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

Answer 33

tis the rt one

Answer 34

lol its big,eh??

Answer 35

but I am just algebraically splitting up the fractions, nothing more.

Answer 36

u will be rt if there werent powers in the Dr

Answer 37

I will be what?

Answer 38

now U have to xpand compare LHS and RHS........good luck with that!!:)

Answer 39

Right=rt

Answer 40

the powers aint matter do they,
\[\frac{1}{(x+1)^2(x-1)^2}=\frac{A}{(x+1)^2}+\frac{B}{(x-1)^2}\]and then I solved for A and B< what is wrong with doing that?

Answer 41

Yes LOL they do

Answer 42

did U read the Article man??

Answer 43

that article from wiki, is a headache, sorry. (just like anything on wiki)

Answer 44

kinda.....but I told u to catch a glimpse of that last point.......It clears everything

Answer 45

cmon.......One more

Answer 46

one more what?

Answer 47

these are gtin interesting!!

Answer 48

question

Answer 49

maybe, I am just trying to practice the method...

Answer 50

the methods actually.

Answer 51

I mean give something more tough if u have got it

Answer 52

dude, I am not a mathemtician like you, I barely learned a u-sub in class.

Answer 53

anyways, I might open a new question.

Answer 54

tnx for helping

Answer 55

sure man ..anytime:)