Question

Calculus- integration 4/(x^6-8 x^3) find partial fraction decomp

Answer 1

I need help finding the decomposition. Please

Answer 2

hello

Answer 3

can you factor the bottom any?

Answer 4

hi freckles..again

Answer 5

the bottom is what has me confused im thinking pull out \[x ^{3}\]

Answer 6

if i did that then it would be \[x^3 (x^2 - 8)\] is that correct?

Answer 7

almost

Answer 8

x^3*x^3=x^(3+3)=x^6 correct?

Answer 9

\[\frac{4}{x^6-8x^3}=\frac{4}{x^3(x^3-8)}=\frac{4}{x^3(x-2)(x^2+2x+4)}\]

Answer 10

do you know how I factor x^3-8 ?

Answer 11

2^3 = 8 so you pull out a factor of x-2

Answer 12

well it was just a difference of cubes

Answer 13

after the x-2 did you use long division again here?

Answer 14

oh

Answer 15

you could if you didn't know the difference of cubes formula

Answer 16

im gonna have to look that up

Answer 17

\[\text{ Difference of cubes } \\ u^3-a^3=(u-a)(u^2+ua+a^2) \\ \text{ Sum of cubes } \\ u^3+a^3=(u+a)(u^2-ua+a^2)\]

Answer 18

the numerator..... would that be A + B + Cx+D ?

Answer 19

Well the thing is we have a linear repeated three times
And we have another linear after that (with no repetition )
and a quadratic

Answer 20

for a linear the numerator should be a constant
for a quadratic the numerator should be a linear

Answer 21

so the x^3 would be Ax+B

Answer 22

\[\frac{4}{x^3(x-2)(x^2+2x+4)} \\ =\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x-2}+\frac{Ex+F}{x^2+2x+4}\]

Answer 23

oh wow

Answer 24

because for the x^3.. we have to show the other factors?

Answer 25

well don't look at x^2 and x^3 as a quadratic and cubic respectively
look at them as linears raise to a power

Answer 26

why doesnt the X^2 and x^3 have Ax+B in the numerator is the degree is higher than 1?

Answer 27

i thought anything over x^1 would need the Ax+B in the numerator. This is what really confuses me

Answer 28

it is because x is linear

Answer 29

the thing inside the power is linear

Answer 30

oh.. ok!

Answer 31

so thhe Ex+F..... that was given the x and 2 coefficients because it can not factor?

Answer 32

linears go over irreducible quadratics

Answer 33

my problem occurs with the numerator and I get confused to whether or not it will need the Ax+B in it

Answer 34

ok

Answer 35

thanks again

Answer 36

\[\text{ pretend we have } \frac{x+1}{x^3}\]
like we aren't go to look at the bottom as a cubic
we are going to look at as a linear cubed

Answer 37

ok..

Answer 38

\[\frac{x+1}{x^3}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}\]

Answer 39

ok

Answer 40

if we had \[\frac{x+1}{(x^2+1)^3}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}+\frac{Ex+F}{(x^2+1)^3}\]

Answer 41

that is because we had a irreducible quadratic cubed

Answer 42

ooohhh I see

Answer 43

i see. and x by itself, no matter the degree...is considered linear

Answer 44

\[\frac{x+1}{x^3(x^2+1)^3}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}+\frac{Hx+J}{(x^2+1)^3}\]

Answer 45

oh my gosh you are soo helpful! I love when your on here

Answer 46

yes

Answer 47

thanks a bunch!

Answer 48

one more example though just for fun
so you can determine a linear from a quadratic or whatever

Answer 49

k

Answer 50

\[\frac{x}{(x-2)^2(x-1)(x^2+1)^2}\]
how would you set this one up

Answer 51

i am slow at the equation thing, so gimme a min to type it in lol

Answer 52

it's cool

Answer 53

\[A/ (x-2) + B/ (x-2)^+ C/(x-1)+ Dx+E/(x^2+1) + Fx+G/ (x^2 +1)^2\]

Answer 54

i get confused when im typing

Answer 55

this is way wrong huh?

Answer 56

I think you just said this
and if you did I'm totally totally proud of you
\[\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x-1}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}\]
you missed the power on the second fraction but I think you meant to put it :)

Answer 57

that plus sign is supposed to be a 2 :(

Answer 58

so i did it right?

Answer 59

yep

Answer 60

awesome! I thank you kindly. Can i Private Message you in the future? You are so helpful to me

Answer 61

sure

Answer 62

great!