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)

- anonymous

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- amistre64

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

- anonymous

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

- anonymous

enter key?

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## More answers

- amistre64

the
key
on
the
keyboard
that
makes
this
happen ;)

- anonymous

you want me to use the enter key more often?

- amistre64

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

- anonymous

but you get radicals

- amistre64

radicals are fine.... right?

- anonymous

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

- amistre64

which other way can you decompose the fraction?

- anonymous

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

- anonymous

technically the quadratic is reducible

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

in general you should, to account for multiples i think

- amistre64

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

- amistre64

right?

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

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

- anonymous

sometimes it is difficult to factor a quadratic, right?

- anonymous

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

- amistre64

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

- anonymous

yes but doing partial fractions with irrational roots is tough

- amistre64

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

- anonymous

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

- anonymous

i was stuck

- anonymous

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

- amistre64

confident? lol ; just stubborn

- anonymous

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

- anonymous

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

- anonymous

theorem time, one sec

- amistre64

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?

- anonymous

yes looks good

- anonymous

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

- anonymous

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

- anonymous

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

- amistre64

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))

- anonymous

MOST calculus books and websites say the quadratic is irreducible though

- anonymous

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

- anonymous

whatever field you have in mine

- anonymous

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

- anonymous

newton, be careful though

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

newtons, so just change y back to x^2

- anonymous

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.

- anonymous

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

- anonymous

correct, i thought you were doing a change of variables

- amistre64

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

- anonymous

amistre, you get medal for persistence :)

- anonymous

but its much easier if we dont split the quadratic

- anonymous

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.

- amistre64

yay!! lol :) can I stop now lol

- anonymous

ok

- anonymous

wait that doesnt work though

- anonymous

I mixed up a minus and plus sign ¬_¬

- anonymous

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

- anonymous

you wont be able to find A and B

- anonymous

I just did?

- anonymous

you used a simpler case though

- anonymous

Multiply it out, plz.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

1 = Ax^2 + Bx62 - 15 A

- anonymous

hmmmm,

- anonymous

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

- anonymous

well this approach works as well

- anonymous

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

- anonymous

right, so come the theorem doesnt work in these websites

- anonymous

as far as repeated roots go

- anonymous

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

- anonymous

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

- anonymous

ahh

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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)

- anonymous

Yes, that.

- anonymous

but its not technically irreducible

- anonymous

i can even go so far as to do this

- anonymous

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

- anonymous

no chat room

- anonymous

sorry, not for you

- anonymous

so this whole irreducible business is unnecessary

- anonymous

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

- anonymous

ok im going to solve an easier one

- anonymous

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

- anonymous

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

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