Question

I need a bit of help with integration

Answer 1

\[\int\limits \frac{ 4x^3+3x^2+2x+1 }{ x^4-1 }\]

Answer 2

*dx

Answer 3

Looks a lot like you want to do some partial fraction decomposition at first, maybe a long hand division at first :-)

Answer 4

\[\int\limits\frac{4x^3+3x^2+2x+1}{x^4-1} \qquad = \qquad \int\limits \frac{4x^3}{x^4-1}dx+\int\limits \frac{3x^2+2x+1}{x^4-1}dx\]

Hmm If we split up the fraction like this, I think we can do a nice easy U-sub on the first integral.

Then for the other one, maybe some factoring, hmm.

Answer 5

Err ya I guess partial fractions would work better :\

Answer 6

Yea, so how would I start this off?

Answer 7

Start by factoring the bottom:
We have the difference of squares,

\[\large x^4-1 \qquad = \qquad (x^2)^2-(1)^2\]Remember how to break down the difference of squares into factors?

Answer 8

(x^2-1)(x^2+1)?

Answer 9

Good good, looks like we can break down that first set of brackets by repeating the rule :)

Answer 10

\[\int\limits \frac{ 4x^3+3x^2+2x+1 }{ (x+1)(x-1)(x^2+1) } dx\]

Answer 11

looks good.
do you understand how to do the initial setup for the partial fractions?

Answer 12

\[\frac{ 4x^3+3x^2+2x+1 }{ (x-1)(x+1)(x^2+1) }=\frac{ A }{ (x-1) }+\frac{ B }{ (x+1) }+\frac{ Cx+D }{ x^2+1 }\]

Answer 13

go ahead, friend. you are good

Answer 14

Yah good job :) looks correct so far.

Answer 15

@zepdrix If he steps up on the right track, give him medal, OK? hihi..

Answer 16

\[4x^3+3x^2+2x+1=A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x^2-1)\]

Answer 17

You might not want to combine those terms on the (Cx+D).
The square will make it harder to solve for your constants.

Well unless you plan on multiplying everything out, then I guess it doesn't matter.
Depends what method you use.

Answer 18

that method confuses me

Answer 19

So if you leave the last term as (Cx+D)(x-1)(x+1),
we can solve for A and B fairly easily,

Let \(\large x=1\), and you can solve for A.
Let \(\large x=-1\), and you can solve for B.

Answer 20

We might still have to multiply it all out to solve for C and D though.. hmm

Answer 21

So 10=4A
A=2.5
----------------------
-2=-4B
B=.5

Answer 22

5/2 and 1/2?
Ya, cool c:

Answer 23

We could probably solve for D by plugging on \(\large x=0\)

Answer 24

I got 1=C
and D=1

Answer 25

cool, sounds correct.

Answer 26

So my only real problem is plugging all the info back into the integration equation..

Answer 27

Ok, it's not as bad as it seems.
We want to plug everything back into this,
\[\large \frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}\]

And then we'll take the integral of that, term by term.

Answer 28

So just plug in the values to the variables?
Sorry about that delay