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

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

enter key?

the
key
on
the
keyboard
that
makes
this
happen ;)

you want me to use the enter key more often?

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

but you get radicals

radicals are fine.... right?

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

which other way can you decompose the fraction?

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

technically the quadratic is reducible

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

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

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

in general you should, to account for multiples i think

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

right?

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

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

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

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

sometimes it is difficult to factor a quadratic, right?

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

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

yes but doing partial fractions with irrational roots is tough

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

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

i was stuck

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

confident? lol ; just stubborn

theorem time, one sec

yes looks good

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

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

MOST calculus books and websites say the quadratic is irreducible though

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

whatever field you have in mine

newton, be careful though

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

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

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

newtons, so just change y back to x^2

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

correct, i thought you were doing a change of variables

amistre, you get medal for persistence :)

but its much easier if we dont split the quadratic

yay!! lol :) can I stop now lol

ok

wait that doesnt work though

I mixed up a minus and plus sign ¬_¬

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

you wont be able to find A and B

I just did?

you used a simpler case though

Multiply it out, plz.

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

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

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

1 = Ax^2 + Bx62 - 15 A

hmmmm,

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

well this approach works as well

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

right, so come the theorem doesnt work in these websites

as far as repeated roots go

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

ahh

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

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

Yes, that.

but its not technically irreducible

i can even go so far as to do this

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

no chat room

sorry, not for you

so this whole irreducible business is unnecessary

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

ok im going to solve an easier one

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

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