Question

Would this question be partial fraction decomposition?

Answer 1

\[\int\limits_{}^{}\frac{ 3x^3-x^2+6x-4 }{ (x^2+1)(x^2+2) }dx\]

Answer 2

you could use partial fractions since the deg of denominator>deg of numerator

Answer 3

Pretty sure that it is, i'm not sure how to break up the bottom so that it remains equivalent
\[\frac{ A }{ (x^2+1) }+\frac{ B }{ (x^2+2) }\] there's some missing right?

Answer 4

those are quadratics
you need a linear numerator Ax+B and Cx+D

Answer 5

In addition to those I listed?

Answer 6

no

Answer 7

you have chosen a constant top for a quadratic denominator

Answer 8

you should have chosen a linear top for a quadratic denominator

Answer 9

\[\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2}\]

Answer 10

oh.
This is going to sound like a silly question..but how do you come up with those on the top?

Answer 11

we want the numerators to be one less than the degree of the denominator

Answer 12

so if you have a linear bottom then you have a constant top

Answer 13

if you have a cubic on bottom then you have a quadratic on top

Answer 14

assuming these quadratics we have on bottom and these cubics if we had cubics on bottom were not-factorable

Answer 15

So if we had like
m/x^4
It would be..
Ax^3/x^4 ?
idk if that example makes sense xD

Answer 16

well no those are repeated linear factors

Answer 17

linear^4

Answer 18

\[\frac{x+1}{(x+1)^4(x^2+1)^2(x^2+2)(x^4+2)^2} \\ \text{ then we would set \it up like this } \\ \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}+\frac{D}{(x+1)^4}+ \\ +\frac{Ex+F}{x^2+1}+\frac{Gx+H}{(x^2+1)^2} \\ + \frac{Ix+J}{x^2+2} \\ +\frac{Kx^3+Lx^2+Nx+P}{x^4+2}+\frac{Qx^3+Rx^2+Sx+T}{(x^4+2)^2} \]
by the way we aren't going to actually find those constants...
this is just a really big example I thought would be nice

Answer 19

yeaah 0.0 woah haha.

Answer 20

oops I didn't mean to chose one where you could cancel a common factor from top and bottom

Answer 21

\[\frac{x+5}{(x+1)^4(x^2+1)^2(x^2+2)(x^4+2)^2} \\ \text{ then we would set \it up like this } \\ \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}+\frac{D}{(x+1)^4}+ \\ +\frac{Ex+F}{x^2+1}+\frac{Gx+H}{(x^2+1)^2} \\ + \frac{Ix+J}{x^2+2} \\ +\frac{Kx^3+Lx^2+Nx+P}{x^4+2}+\frac{Qx^3+Rx^2+Sx+T}{(x^4+2)^2}\]
there now I'm happy

Answer 22

So when we set mine equal to the numerator..how do we do the right side?
3x^3-x^2+6x-4= ..?
Since we have Ax+B all over the same den

Answer 23

haha it's beautiful xD

Answer 24

same denominator?

Answer 25

\[\frac{ Ax+B }{ (x^2+1) }\]

Answer 26

Would it be
(Ax+B)(x^2+2) ?

Answer 27

\[\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+2} \\ \frac{(Ax+B)(x^2+2)+(Cx+D)(x^2+1)}{(x^2+1)(x^2+2)}\]

Answer 28

just multiply first fraction by (x^2+2)/(x^2+2)
and the second one by (x^2+1)/(x^2+1)
so you will have same denominator and are able to combine fractions

Answer 29

so you have \[3x^3-x^2+6x-4=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)\]

Answer 30

oh excellent :)
now to solve!

Answer 31

you try to use heaviside
like for example pluggin in i gives
\[3i^3-i^2+6i-4=(Ai+B)(i^2+2) \\ -3i+1+6i-4=(Ai+B)1 \\ 3i-3=Ai+B \\ \implies A=3 \text{ and } B=-3\]
you try putting in sqrt(2)i if you want

Answer 32

or you could multiply everything out and regroup the items

Answer 33

and then compare both sides to find A and B, and C and D

Answer 34

can't we just plug in values for x?

Answer 35

But I guess values like x=0 wouldn't work at all

Answer 36

you could it just takes longer

Answer 37

like I can just choose 2 values and be done with it

but if you choose other values... I think since we have 4 unknowns we are going to have to pick 4 numbers to plug in

Answer 38

Oh, I see.
and I would be choosing those two values for what?
lol these I didn't get at all -.-

Answer 39

did you understand how I got A and B above?

Answer 40

not really xD
nor why you dropped te (Cx+D)(x^2+1) ?

Answer 41

\[3x^3-x^2+6x-4=(Ax+B)(x^2+2)+(Cx+D)(x^2+1) \\ \text{ I know I could choose } i \text{ or } -i \text{ and that one } \\ \text{ term will disappear } \]

Answer 42

why because x^2+1 is (x-i)(x+i)
puttin in i or -i will make one of those factors 0

Answer 43

0*(Cx+D) is 0

Answer 44

\[x=i \\ 3x^3-x^2+6x-4=(Ax+B)(x^2+2)+(Cx+D)(x^2+1) \\ 3(i)^3-i^2+6(i)-4=(Ai+B)(i^2+2)+(Ci+D)(i^2+1) \\ \\ \text{ recall } i^2=-1 \text{ and } i^3=-i \\ \\ -3i-(-1)+6i-4=(Ai+B)(-1+2)+(Ci+D)(-1+1) \\ -3i+1+6i-4=(Ai+B)(1)+(Ci+D)(0) \\ 3i-3=Ai+B\]

Answer 45

then compare imaginary parts and real parts
A=3 and B=-3

Answer 46

\[x^2+2=(x-\sqrt{2}i)(x+\sqrt{2}i)\]
so choosing either sqrt(2)i or -sqrt(2)i will work to find C and D

Answer 47

just like above those we didn't need to do both options
just one

Answer 48

\[3(\sqrt{2}i)^3-(\sqrt{2}i)^2+6(\sqrt{2}i)-4=0+(C \sqrt{2 } i+D)((\sqrt{2}i)^2+1)\]

Answer 49

this isn't as bad as it looks

Answer 50

So..in "comparing imaginary and real parts"
because the 3 and the A have i beside them, we know those go together?
likewise with the B and 3 having a lack of i?

Answer 51

yep

Answer 52

Ai+B=Ci+D if this equality holds assuming A,B,C, and D are real
then A=C and B=D

Answer 53

if you don't like the way that cube term looks I can show you it isn't too bad if you need me to

Answer 54

\[(\sqrt{2} i)^3=(\sqrt{2})^3 i^3=(\sqrt{2})^2 (\sqrt{2})^1(-i)=2 \sqrt{2}(-i)=-2i \sqrt{2}\]
does this make sense to you?

Answer 55

It's mostly the i that is throwing me off.

Answer 56

ok you don't have to plug in imaginary numbers if you aren't comfortable with them

Answer 57

plug in x=0,1,-1, and...

Answer 58

I guess 2

Answer 59

or if you don't want to do heaviside
expand and regroup and compare both sides

Answer 60

\[3x^3-x^2+6x-4=(Ax+B)(x^2+2)+(Cx+D)(x^2+1) \\\ 3x^3-x^2+6x-4 \\ =Ax(x^2+2)+B(x^2+2)+Cx(x^2+1)+D(x^2+1) \\ =Ax^3+2Ax+Bx^2+2B+Cx^3+Cx+Dx^2+1 \\ =(A+C)x^3+(B+D)x^2+(A+C)x+(2B+D)\]

Answer 61

with x=0 we get
-4=2B+D

Answer 62

that is right

Answer 63

select 3 more values

Answer 64

to plug in

Answer 65

x=1
4=3A+3B+2C+2D

Answer 66

that looks right too

Answer 67

x=-1
-4=-3A-3B-2C-2D

Answer 68

oops the left is
-12

Answer 69

I have to do it on paper I'm getting -14

Answer 70

mmm still gettinng -12 o.o

Answer 71

\[3x^3-x^2+6x-4=(Ax+B)(x^2+2)+(Cx+D)(x^2+1) \\ 3(-1)^3-(-1)^2+6(-1)-4=(A \cdot -1+B)((-1)^2+2)+(C \cdot -1+D)((-1)^2+1) \\ \\ 3(-1)-1-6-4=(-A+B)(3)+(-C+D)(2) \\ -3-1-6-4=-3A+3B-2C+2D \\ -4-6-4=-3A+3B-2C+2D \\ -14=-3A+3B-2C+2D\]
I still could be doing something wrong but this is what I have

Answer 72

eerm. idk you're probably corect lol

Answer 73

http://www.wolframalpha.com/input/?i=f(x)%3D3x%5E3-x%5E2%2B6x-4;+evaluate+f(-1)

Answer 74

i think your mistake is occuring the square term

Answer 75

maybe you are for some reason replacing minus (-1)^2 with + (1)^2 ?

Answer 76

when it should be minus (1)^2

Answer 77

or minus 1

Answer 78

maybe o.0

Answer 79

you don't sound convinced yet :p

Answer 80

can i see your operations please

Answer 81

like how you are working it

Answer 82

Yes, just a sec :) dinner @.@

Answer 83

@freckles sorry just found out about a hw due tomorrow so i'm working on that D: