Question

∫(dx)/[x/(x^2-1)^2],solve using partial fraction

Answer 1

can jus tell me the final ans.

Answer 2

jimmy.r u solving the qn?

Answer 3

i'm trying - i'm a bit rusty at these - i,m consulting the text book!!

Answer 4

so might be a while

Answer 5

ok.I'll wait

Answer 6

jimmy, r u working out the solution?

Answer 7

ya

Answer 8

how?

Answer 9

i've got 4 fractions so 4 unknowns - the algebra is horrendous - there must be an easier way to do this - ill try wolframappha

Answer 10

ok.

Answer 11

but what is ur final ans?

Answer 12

i've split it up into four fractions so 4 unknowns - very messy - there must be an easier way - i'm checking it out on wolfram alpha

Answer 13

my messages arent getting thru

Answer 14

yup,wat jimmy?

Answer 15

ok. anyone with final ans?

Answer 16

i'm trying wolframalpha but its coming up with error messages all the time

Answer 17

ok

Answer 18

is this your integral?

∫(dx)/[x/(x^2-1)^2]

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

(Why doesn't thins thing like the negative sign??)

Answer 19

yes,right

Answer 20

math - i gotta go - sorry i couldn't help

Answer 21

ok

Answer 22

so it can be written \[\int\frac{(x^2-1)^2}{x}dx\]

Answer 23

no.

Answer 24

numerator is on e denominator, that means 1 is on e numerator

Answer 25

denominator has[x[x^{2}-1]^{2}\]

Answer 26

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

Answer 27

yes ur right!

Answer 28

ok..give me a min and I'll give you an answer

Answer 29

ok.

Answer 30

this is from my ti-nspire

\[\frac{-\left[(x^2-1)\ln|x^2-1|-2(x^2-1)\ln|x|+1\right]}{2(x^2-1)}\]

Answer 31

or from wolfram

http://www.wolframalpha.com/input/?i=integral [1%2F%28x*%28x^2-1%29^2%29]+&t=ietb01

Answer 32

is this ur ans?

Answer 33

http://alturl.com/vfbsr

Answer 34

my nspire combines everything..the wolfram answer is nicer looking

Answer 35

ok. thanks