Question

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

Answer 1

simple, split \[x^{2}-1 = (x+1)(x-1)\]

Answer 2

y do u want to do this partial fractions jus substitute x^2 = t nd ull get ur ans. in seconds

Answer 3

But myabe they are asked to solve using partial fractions!

Answer 4

rather substiture x^2-1 = t will be more easy

Answer 5

@mashy i still havnt studied partial fractions ...
i will b completing dat till tomorow den i Can Answer this question :)

Answer 6

yes manjotpal bhaiya is right mashy

Answer 7

(x^2-1) whole thing squared again kushashwa

Answer 8

i didn't understand what u want to say mathstina

Answer 9

if u r a beginner in partial fraction then pls follow the following LINKS

Answer 10

http://en.wikipedia.org/wiki/Partial_fraction
http://www.purplemath.com/modules/partfrac.htm
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html
http://www.mathsrevision.net/alevel/pages.php?page=94
these links would help u by examples and everything u want

Answer 11

if this does not help or if this is not that u wanted then tell

Answer 12

no i have books that deal to a great lot extent in Caculus...

but ill folloe the above links too . thnx :)

Answer 13

no problem

Answer 14

but i was mainly saying to the asker of this question but u can also view it bhaiya

Answer 15

joe, can u solve?

Answer 16

im a little confused as to what the function looks like...is it this?
\[\frac{dx}{\frac{x}{x^{2}-1}}\]

Answer 17

or this:
\[\frac{x}{x^{2}-1}dx\]

Answer 18

yes,first one

Answer 19

then you cant really apply partial fractions because:
\[\frac{1}{\frac{x}{x^{2}-1}}dx = \frac{x^2-1}{x}dx\]
then you would just divide it out to get x-(1/x), and the integral would be x^2/2-ln(x)

Answer 20

the second one looks like a good candidate for partial fractions.

Answer 21

yes,first one

Answer 22

thanks for ur help. i need to sign out nw.