Question

integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^2-15)) should i do partial fraction 1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)

Answer 1

oy... if you hit the enter key it spreads this out to where it is more readable.... just a thought

Answer 2

the x^2 - 15 is reducible , but im not sure if youre supposed to use that theorem

Answer 3

enter key?

Answer 4

the

key

on

the

keyboard

that

makes

this

happen ;)

Answer 5

you want me to use the enter key more often?

Answer 6

yes, difference of squares that part and you should have 3 sets to figure out

Answer 7

but you get radicals

Answer 8

radicals are fine.... right?

Answer 9

are they ? how come the other way works just as well?

Answer 10

which other way can you decompose the fraction?

Answer 11

1/((x^2*(x^2-15)) = A / x + B/x^2 + (Cx+d) / ( x^2 - 15)

Answer 12

technically the quadratic is reducible

Answer 13

it could be a fluke if it does work out....

Answer 14

or it could be a possibility for solutions that i am unaware of as yet :)

Answer 15

no it isnt a fluke, in general one need not reduce a quadratic

Answer 16

in general you should, to account for multiples i think

Answer 17

if it aint got multiples, then you should be fine i spose

Answer 18

right?

Answer 19

yes it usually makes it easier to reduce the quadratic, except in the case where there are radical linear factors

Answer 20

it can have multiples, theres no problem as far as i can see

Answer 21

without doing a ton of them to generalize it, id have to throw it in the hmmmmm box lol

Answer 22

say integral ( 1 / (x(x+1)^2) = A/x + (Bx+C) / (x+1)^2

Answer 23

its a little weird, might be easier if i wrote
1/( x ( x^2+2x+1)) = A/x + (Bx+c) / (x^2 + 2x + 1)

Answer 24

sometimes it is difficult to factor a quadratic, right?

Answer 25

ok , i should get into more accurate terms. sometimes it is reducible over reals, not rationals

Answer 26

factoring a quadratic is fairly easy; might get complicated solutions, but farily easy to do

Answer 27

yes but doing partial fractions with irrational roots is tough

Answer 28

and i have heard that you can use the complex solutions as well with no ill effects on the decomp

Answer 29

ok , so youre so confident, try to do parfractions on 1 / ( x^2(x^2-15) , give it a go

Answer 30

i was stuck

Answer 31

1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15)

Answer 32

confident? lol ; just stubborn

Answer 33

it can even get stranger, this is an easy irrational solution problem, you can have say A/
( x - sqrt 3/2 ) for instance

Answer 34

this is why calculus in textbooks is generally contrived , or idealized problems. in the real world you have these irrational roots, etc

Answer 35

theorem time, one sec

Answer 36

1 a b c d
----------- = --- + ----- + --------- + ---------
x^2(x^2-15) x x^2 x-sqrt(15) x+sqrt(15)

is the 'proper' set up right?

Answer 37

yes looks good

Answer 38

Personally, I'd leave the last fraction's denominator without doing difference of two squares.

Answer 39

1 / [(x-a)^n (x^2+bx+c)^n] = A1/(x-a) + A2/ ( x-a)^2 + ... An/(x-a)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

Answer 40

so the question is, do the quadratics have to be irreducible, i dont think so

Answer 41

1 a b c d
----------- = --- + ----- + --------- + ---------
x^2(x^2-15) x x^2 x-sqrt(15) x+sqrt(15)

a(x^2)(x-sqrt(15))(x+sqrt(15)))

b(x)(x-sqrt(15))(x+sqrt(15)))

c(x)( x^2)(x+sqrt(15))

d(x)(x^2)(x-sqrt(15))

Answer 42

MOST calculus books and websites say the quadratic is irreducible though

Answer 43

irreducible over the reals i imagine. wikipedia is better, they say irreducible over the field

Answer 44

whatever field you have in mine

Answer 45

For this particular problem:

\[\frac{1}{y(y-15)} \equiv \frac{1}{15(y-15)} - \frac{1}{15y}\]
\[\text{Let } y = x^2\] ...

Answer 46

newton, be careful though

Answer 47

if you are substituting a variable, you will need to do 2x dx = dy

Answer 48

so you might want to just change it back. thats an identity

Answer 49

i should write a book on calculus pitfalls, since i keep finding them :)

Answer 50

newtons, so just change y back to x^2

Answer 51

Sorry, in case it wasn't clear, it leads to:

\[\frac{1}{x^2(x^2-15)} \equiv \frac{1}{15(x^2-15)} + \frac{1}{15x^2} \]

I would integrate this.

Answer 52

otherwise you have 2x dx = dy , dx = dy / ( 2x ) = dy / ( 2 sqrt y )

Answer 53

correct, i thought you were doing a change of variables

Answer 54

1 =

a(x^2)(x-sqrt(15))(x+sqrt(15)))

+ b(x)(x-sqrt(15))(x+sqrt(15)))

+ c(x)( x^2)(x+sqrt(15))

+d(x)(x^2)(x-sqrt(15))

when x = sqrt(15) we get

1 = c(sqrt(15))(15)(2sqrt(15))

1 = c(15)(2)(15)

1 = c(225)(2) = c450

c = 1/450

Answer 55

amistre, you get medal for persistence :)

Answer 56

but its much easier if we dont split the quadratic

Answer 57

I was only doing that to show that you don't need a 1/x fraction and 1/x^2 , as x^2 is the only term. I wasn't integrating with a different variable.

Answer 58

yay!! lol :) can I stop now lol

Answer 59

ok

Answer 60

wait that doesnt work though

Answer 61

I mixed up a minus and plus sign ¬_¬

Answer 62

its incorrect to do 1 / ( x^2 (x^2-15 ) = A / x^2 + b / (x^2-15) , thats incorrect

Answer 63

you wont be able to find A and B

Answer 64

I just did?

Answer 65

you used a simpler case though

Answer 66

Multiply it out, plz.

Answer 67

1 = A(x^2 - 15) + B x^2

Answer 68

, 1 = Ax^2 + bx^2 - 15

Answer 69

(the + the last fraction should be minus, as in the first time, typo)

Answer 70

1 = Ax^2 + Bx62 - 15 A

Answer 71

hmmmm,

Answer 72

but the book says it should be A/ x + B/x^2 + C x+d / (x^2 - 15)

Answer 73

well this approach works as well

Answer 74

You can do it if you want, but you'll find A = 0, so it doesn't really matter.

Answer 75

right, so come the theorem doesnt work in these websites

Answer 76

as far as repeated roots go

Answer 77

for instance you didnt use CX+D, i think its because of that y = x^2 business,

Answer 78

There are only terms in x^2 in this case, it wouldn't always work. The general method is probably more ... general.

Answer 79

ahh

Answer 80

so my question, my original question, hmmmm, do the quadratics have to be irreducible in the general case

Answer 81

ok for instance, 1 / (( x-2)^2 ( x^2-15))

Answer 82

1 / (( x-2)^2 ( x^2-15)) = A / (x-2) + b / (x-2)^2 + (cx+d) / (x^2 - 15 ) ?

Answer 83

I'd say it's probably easier to leave it as the irreducible (without using difference of two squares) quadratic with Cx + d. I guess you could split up this factor, but it's unnecessary (as far as I know)

Answer 84

Yes, that.

Answer 85

but its not technically irreducible

Answer 86

i can even go so far as to do this

Answer 87

1 / (( x-2)^2 ( x^2-15)) = (Ax+b) / (x-2)^2 + (cx+d) / (x^2 - 15 )

Answer 88

no chat room

Answer 89

sorry, not for you

Answer 90

so this whole irreducible business is unnecessary

Answer 91

1 = a*x^3-15*a*x+b*x^2-15*b+c*x^3-4*c*x^2+4*c*x+d*x^2-4*d*x+4*d

Answer 92

ok im going to solve an easier one

Answer 93

int 1/ ((x-2)(x^2-15))

Answer 94

1/ ((x-2)(x^2-15)) = A/ (x-2) + (Bx+c) / ( x^ 2 - 15)