anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
oy... if you hit the enter key it spreads this out to where it is more readable.... just a thought
anonymous
  • anonymous
the x^2 - 15 is reducible , but im not sure if youre supposed to use that theorem
anonymous
  • anonymous
enter key?

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amistre64
  • amistre64
the key on the keyboard that makes this happen ;)
anonymous
  • anonymous
you want me to use the enter key more often?
amistre64
  • amistre64
yes, difference of squares that part and you should have 3 sets to figure out
anonymous
  • anonymous
but you get radicals
amistre64
  • amistre64
radicals are fine.... right?
anonymous
  • anonymous
are they ? how come the other way works just as well?
amistre64
  • amistre64
which other way can you decompose the fraction?
anonymous
  • anonymous
1/((x^2*(x^2-15)) = A / x + B/x^2 + (Cx+d) / ( x^2 - 15)
anonymous
  • anonymous
technically the quadratic is reducible
amistre64
  • amistre64
it could be a fluke if it does work out....
amistre64
  • amistre64
or it could be a possibility for solutions that i am unaware of as yet :)
anonymous
  • anonymous
no it isnt a fluke, in general one need not reduce a quadratic
amistre64
  • amistre64
in general you should, to account for multiples i think
amistre64
  • amistre64
if it aint got multiples, then you should be fine i spose
amistre64
  • amistre64
right?
anonymous
  • anonymous
yes it usually makes it easier to reduce the quadratic, except in the case where there are radical linear factors
anonymous
  • anonymous
it can have multiples, theres no problem as far as i can see
amistre64
  • amistre64
without doing a ton of them to generalize it, id have to throw it in the hmmmmm box lol
anonymous
  • anonymous
say integral ( 1 / (x(x+1)^2) = A/x + (Bx+C) / (x+1)^2
anonymous
  • 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
  • anonymous
sometimes it is difficult to factor a quadratic, right?
anonymous
  • anonymous
ok , i should get into more accurate terms. sometimes it is reducible over reals, not rationals
amistre64
  • amistre64
factoring a quadratic is fairly easy; might get complicated solutions, but farily easy to do
anonymous
  • anonymous
yes but doing partial fractions with irrational roots is tough
amistre64
  • amistre64
and i have heard that you can use the complex solutions as well with no ill effects on the decomp
anonymous
  • anonymous
ok , so youre so confident, try to do parfractions on 1 / ( x^2(x^2-15) , give it a go
anonymous
  • anonymous
i was stuck
anonymous
  • anonymous
1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15)
amistre64
  • amistre64
confident? lol ; just stubborn
anonymous
  • 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
  • anonymous
this is why calculus in textbooks is generally contrived , or idealized problems. in the real world you have these irrational roots, etc
anonymous
  • anonymous
theorem time, one sec
amistre64
  • 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
  • anonymous
yes looks good
anonymous
  • anonymous
Personally, I'd leave the last fraction's denominator without doing difference of two squares.
anonymous
  • 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
  • anonymous
so the question is, do the quadratics have to be irreducible, i dont think so
amistre64
  • 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
  • anonymous
MOST calculus books and websites say the quadratic is irreducible though
anonymous
  • anonymous
irreducible over the reals i imagine. wikipedia is better, they say irreducible over the field
anonymous
  • anonymous
whatever field you have in mine
anonymous
  • 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
  • anonymous
newton, be careful though
anonymous
  • anonymous
if you are substituting a variable, you will need to do 2x dx = dy
anonymous
  • anonymous
so you might want to just change it back. thats an identity
anonymous
  • anonymous
i should write a book on calculus pitfalls, since i keep finding them :)
anonymous
  • anonymous
newtons, so just change y back to x^2
anonymous
  • 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
  • anonymous
otherwise you have 2x dx = dy , dx = dy / ( 2x ) = dy / ( 2 sqrt y )
anonymous
  • anonymous
correct, i thought you were doing a change of variables
amistre64
  • 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
  • anonymous
amistre, you get medal for persistence :)
anonymous
  • anonymous
but its much easier if we dont split the quadratic
anonymous
  • 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
  • amistre64
yay!! lol :) can I stop now lol
anonymous
  • anonymous
ok
anonymous
  • anonymous
wait that doesnt work though
anonymous
  • anonymous
I mixed up a minus and plus sign ¬_¬
anonymous
  • anonymous
its incorrect to do 1 / ( x^2 (x^2-15 ) = A / x^2 + b / (x^2-15) , thats incorrect
anonymous
  • anonymous
you wont be able to find A and B
anonymous
  • anonymous
I just did?
anonymous
  • anonymous
you used a simpler case though
anonymous
  • anonymous
Multiply it out, plz.
anonymous
  • anonymous
1 = A(x^2 - 15) + B x^2
anonymous
  • anonymous
, 1 = Ax^2 + bx^2 - 15
anonymous
  • anonymous
(the + the last fraction should be minus, as in the first time, typo)
anonymous
  • anonymous
1 = Ax^2 + Bx62 - 15 A
anonymous
  • anonymous
hmmmm,
anonymous
  • anonymous
but the book says it should be A/ x + B/x^2 + C x+d / (x^2 - 15)
anonymous
  • anonymous
well this approach works as well
anonymous
  • anonymous
You can do it if you want, but you'll find A = 0, so it doesn't really matter.
anonymous
  • anonymous
right, so come the theorem doesnt work in these websites
anonymous
  • anonymous
as far as repeated roots go
anonymous
  • anonymous
for instance you didnt use CX+D, i think its because of that y = x^2 business,
anonymous
  • anonymous
There are only terms in x^2 in this case, it wouldn't always work. The general method is probably more ... general.
anonymous
  • anonymous
ahh
anonymous
  • anonymous
so my question, my original question, hmmmm, do the quadratics have to be irreducible in the general case
anonymous
  • anonymous
ok for instance, 1 / (( x-2)^2 ( x^2-15))
anonymous
  • anonymous
1 / (( x-2)^2 ( x^2-15)) = A / (x-2) + b / (x-2)^2 + (cx+d) / (x^2 - 15 ) ?
anonymous
  • 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
  • anonymous
Yes, that.
anonymous
  • anonymous
but its not technically irreducible
anonymous
  • anonymous
i can even go so far as to do this
anonymous
  • anonymous
1 / (( x-2)^2 ( x^2-15)) = (Ax+b) / (x-2)^2 + (cx+d) / (x^2 - 15 )
anonymous
  • anonymous
no chat room
anonymous
  • anonymous
sorry, not for you
anonymous
  • anonymous
so this whole irreducible business is unnecessary
anonymous
  • 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
  • anonymous
ok im going to solve an easier one
anonymous
  • anonymous
int 1/ ((x-2)(x^2-15))
anonymous
  • anonymous
1/ ((x-2)(x^2-15)) = A/ (x-2) + (Bx+c) / ( x^ 2 - 15)

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