Question

Need help evaluating this integral.. I know the answer, just not how to get to it

Answer 1

\[\int\limits_{?}^{?}x^2 /(x^4-1)dx\]

Answer 2

did you try integration by parts?

Answer 3

Yes that did not help..

Answer 4

how far did you get?
did you get this far:
\[\frac{x^2}{x^4-1}=\frac{\frac{1}{4}}{x-1}+\frac{\frac{-1}{4}}{x+1}+\frac{\frac{1}{2}}{x^2+1} ?\]

Answer 5

Or is that the problem getting there?

Answer 6

yeah the problem is getting there..

Answer 7

\[x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)\]
First factor the bottom as much as you can over the rationails

Answer 8

rationals *

Answer 9

alright that makes sense so far

Answer 10

the linear factors we will do constant/the linear factor
the non linear we will do (Constant*x+constant)/the non linear factor

Answer 11

\[\frac{x^2}{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}\]

Answer 12

So these are equal expressions...

Answer 13

Yeah the Cx+D was mostly throwing me off

Answer 14

That means if we combine the fractions on the right hand side then the top should be the same as the top as the left and bottom will be the same as the bottom on the other side

Answer 15

\[\frac{x^2}{x^4-1}=\frac{A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)}{(x-1)(x+1)(x^2+1)}\]

Answer 16

\[\frac{x^2}{x^4-1}=\frac{A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)(x+1)}{x^4-1} \]
The bottoms are the same
The tops should be the same but we need to find A,B,C,D such that it is

Answer 17

Alright I think I got it now. So I get D to=1/2, 1/4 and b=-1/4 then i just put bcak into equation, take the integral and im done?

Answer 18

\[1 \cdot x^2+A(x^3+x+x^2+1)+B(x^3+x-x^2-1)+(Cx+D)(x^2-1)\]

Answer 19

what about A?
C=0 by the way

Answer 20

oh yes are you saying A is 1/4?

Answer 21

yes

Answer 22

right! :)

Answer 23

While your here do you mind telling me how i did this one wrong.. \[\int\limits_{?}^{?} y^4 +y^2-1/(y^3+y)] I went through and got y+ 1/y^3+y which leaves me at a dead end?

Answer 24

I'm having trouble reading it...
one sec

Answer 25

sorry its \[\int\limits_{?}^{?} y^4+y^2-1/(y^3+y)\]

Answer 26

\[\int\limits\limits_{?}^{?}\frac{ y^4 +y^2-1}{y^3+y} dy\]

Answer 27

yes

Answer 28

I do long divison to get y+ 1/y^3+y