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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anonymous
  • anonymous
why don't you type in math. will help in viewing easily.
anonymous
  • anonymous
the second one is the right way. How do you hope getting radicals will help the partial fraction??(Indeed, we use partial fractions for simplification!!!)
anonymous
  • anonymous
hello

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anonymous
  • anonymous
whats wrong with partial fractions with radicals?
anonymous
  • anonymous
i dont see any problem with the first way
anonymous
  • anonymous
no problems but we avoid it as calculations get messy.
anonymous
  • anonymous
my computer does the partial fractions integral 1/(x^2 (x^2-15)) = -1/(15x^2) + 1/(15(x^2-15))
anonymous
  • anonymous
yah...that "seems" easy :) you can always take x^2=y during partial fracs.. then u get 1/(y(y-15)) which can be easily split into fracs.
anonymous
  • anonymous
that doesnt disobey partial fractions rule
anonymous
  • anonymous
i thought it was , it had to be linear factors and irreducibles
anonymous
  • anonymous
like A/x + B/x^2 + C / (x-sqrt 15) + D/(x+sqrt 15)
anonymous
  • anonymous
See...there are no so-called rules...at least not for me :) if we take radicals the math is correct but its unconventional
anonymous
  • anonymous
cx+d is supposed to be only for irreducibles, i thought
anonymous
  • anonymous
like integral 1/x(x^2+1) = A/x + bx+c / x^2 + 1
anonymous
  • anonymous
so youre saying 1/x(x^2-1) = A/x + (bx+c) /( x^2 - 1)
anonymous
  • anonymous
not only for irreducibles,
anonymous
  • anonymous
im pretty sure thats what the theorem states
anonymous
  • anonymous
maybe the theorem u saw said that we cx+d for irreducibles but it did not say that we cant use cx+d for non-irreducibles!!
anonymous
  • anonymous
we use cx+d for irreducibles but we can also use it like (bx+c)/(x^2-1). Here the math is correct. But splitting like A/(x+1)+B/(x-1) is more advantageous, as it gives a good form whose integral is known.
anonymous
  • anonymous
for example, this partial fractions fails 1/(x^2 (x-5) ) = A/x^2 + B/(x-5) , this fails
anonymous
  • anonymous
oh because of the A/x + B/x^2
anonymous
  • anonymous
yes
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
ok so 1/[(x-a)^n (x^2 +bx +c)^n] = k1/(x-a) + k2/(x-a)^2 + ... kn/(x-a)^n + m1x+p1/(x^2 +bx +c) +...
anonymous
  • anonymous
your upto kn is ok. but from m1,p1,...its getting complicated.
anonymous
  • anonymous
yeah im running out of letters
anonymous
  • anonymous
no. there will be cx+d term above and the denominator's power will increase from 1 to n
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... +
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... +(Bnx +Cn) / (x^2 +bx +c)^n
anonymous
  • anonymous
your c1 and b1 terms are wrong
anonymous
  • anonymous
why?
anonymous
  • anonymous
there won't be any b1x ... also instead of c1 write it as b1x+c1
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... +(Bnx +Cn) / (x^2 +bx +c)^n
anonymous
  • anonymous
yups. thats correct....:)
anonymous
  • anonymous
so thats fine?
anonymous
  • anonymous
oh i left out the parenthesees
anonymous
  • anonymous
ok so thats fine for partial fractions , and x^2 +bx + c doesnt necessarily have to be irreducible,
anonymous
  • anonymous
see..make sure that you know that WE USE cx+d in numerator FOR IRREDUCIBLES (in other words polynomials have only complex roots). WE COULD use cx+d also above REDUCIBLES since it is mathematically correct but don't use it as one of the partial fractions is not in its lowest degrees (i.e. cx+d over a reducible is NOT recommended) But if such a use simplifies a problem at hand then why not use it? Understand me?
anonymous
  • anonymous
a reducible quadratic you mean
anonymous
  • anonymous
reducible into linear factors
anonymous
  • anonymous
yups yups a QUADRATIC can only have cx+d above it. yes reducible into linear REAL factors. Nobody wants to have complex factors here!
anonymous
  • anonymous
ok, so then we could do 1/(x^2 (x^2 - 15)) , we could reduce it
anonymous
  • anonymous
im doing 1 / ( x ( x^2 + 2x + 1) ) the latter is reducible, but im not going to . = AX+b / ( x^2 + 2x + 1) + C / x
anonymous
  • anonymous
Mathematically permissible. But not recommended.
anonymous
  • anonymous
ok lets try this one 1/ ( x (x^2+2x+1)) integrate that using parfrac
anonymous
  • anonymous
int 1/ ( x (x^2+2x+1))= (AX+b) / ( x^2 + 2x + 1) + C / x
anonymous
  • anonymous
in part fracs.. its (1/x)+((-x-2)/(x^2+2x+1)) the first term gives log|x| and the second make substitution x^2+2x=y
anonymous
  • anonymous
right, or we could have factored x^2 + 2x + 1
anonymous
  • anonymous
yes...
anonymous
  • anonymous
wait thats wrong
anonymous
  • anonymous
whats that substitution ?
anonymous
  • anonymous
yupss....the subs wrong...
anonymous
  • anonymous
we can do it however
anonymous
  • anonymous
oh by the way, partial fraction doesnt help us with say 1/ (x^(3/2) (x-15))
anonymous
  • anonymous
no.. since the denominator is not a POLYNOMIAL
anonymous
  • anonymous
, ahh
anonymous
  • anonymous
both numer. and denom. must be polynomials
anonymous
  • anonymous
wait is that a precondition?
anonymous
  • anonymous
you could have 1/(x*(x^2-sqrt 3)) for example
anonymous
  • anonymous
YES partial fractions can help ONLY TO RATIONAL FUNCTIONS i.e. both numeratoer and denominator polynomials with denominator not being the zero polynomial
anonymous
  • anonymous
ok
anonymous
  • anonymous
just clearing that up
anonymous
  • anonymous
by the way what's your age?
anonymous
  • anonymous
25
anonymous
  • anonymous
why do i sound dumb
anonymous
  • anonymous
hahha..joking? i hope u understood.. ask for doubts...
anonymous
  • anonymous
i am 17
anonymous
  • anonymous
oh
anonymous
  • anonymous
jeez, i thought you were older. you are wise beyond your years :)
anonymous
  • anonymous
are you a mathematician? or math-related worker? u named urself cantorset...
anonymous
  • anonymous
yes im a math tutor
anonymous
  • anonymous
for a living, i was stumped because the student didnt want to reduce a quadratic
anonymous
  • anonymous
and i thought, hey the theorem says so and so
anonymous
  • anonymous
it was reducible into irrational factors
anonymous
  • anonymous
the question u asked me? - yes it can be solved both ways
anonymous
  • anonymous
see this problem we just did its a lot easier if we reduce the quadratic
anonymous
  • anonymous
integral (1/(x(x^2+2x+1))
anonymous
  • anonymous
yes its lot easier if we reduce quad... but i argued just because i wanted to show that cx+d hold good for reducible quadratics also..
anonymous
  • anonymous
right so i guess it depends on the problem
anonymous
  • anonymous
well its good to know the power of this theorem
anonymous
  • anonymous
which area of math are you specialized in?
anonymous
  • anonymous
i tutor mostly calculus ,
anonymous
  • anonymous
wow..thats awesome..i love calculus..but know little..struggling still with single variable calculus
anonymous
  • anonymous
youre amazing :)
anonymous
  • anonymous
well if you need help, give me a buzz
anonymous
  • anonymous
ok..thank you.
anonymous
  • anonymous
so yeah, this one is tougher to integrate , without reducing the quadratic
anonymous
  • anonymous
integral (1/x)+((-x-2)/(x^2+2x+1))
anonymous
  • anonymous
yups.
anonymous
  • anonymous
so the original question integral 1/(x^2 (x^2-15))
anonymous
  • anonymous
then reducing the quadratic is worse, lol
anonymous
  • anonymous
x^2=y, integr=(1/15)log|(y-15)/y|+constant
anonymous
  • anonymous
then you need 2x dx = dy
anonymous
  • anonymous
oh its not so simple ,
anonymous
  • anonymous
i.e. integral=\[\frac{1}{15}\ln \left| \frac{x^2-15}{x^2} \right|+constant\]
anonymous
  • anonymous
when you make a substitution, you have to do dy = 2x dx
anonymous
  • anonymous
SORRY!! out of my mind......u r right..
anonymous
  • anonymous
then we have another problem
anonymous
  • anonymous
we want dx by itself
anonymous
  • anonymous
dy/(2x) = dx
anonymous
  • anonymous
so we have dx = dy / (2sqrt y))
anonymous
  • anonymous
ok, lets not do the substitution then
anonymous
  • anonymous
\[\frac{1}{15x}+\frac{1}{30\sqrt{15}}\ln \left| \frac{x-\sqrt{15}}{x+\sqrt{15}} \right|+constant\]
anonymous
  • anonymous
i think this is the answer.
anonymous
  • anonymous
let me check
anonymous
  • anonymous
i use maple
anonymous
  • anonymous
1/(15*x)-(1/225)*sqrt(15)*arctanh((1/15)*x*sqrt(15))
anonymous
  • anonymous
must be the same..
anonymous
  • anonymous
oh i messed up
anonymous
  • anonymous
so what did you use ?
anonymous
  • anonymous
\[\int\limits_{}^{}\frac{dx}{x^2-15}=\frac{1}{\sqrt{15}} \ln \left| \frac{x-\sqrt{15}}{x+\sqrt{15}} \right| +C\] i used this identity
anonymous
  • anonymous
but theres an x^2 in front
anonymous
  • anonymous
that is easy!!
anonymous
  • anonymous
\[\int\limits_{?}^{?}1/(x^2(x^2-15))\]
anonymous
  • anonymous
how do you do fraction in latex?
anonymous
  • anonymous
frac{2}{3}=2/3
anonymous
  • anonymous
1/(x^2(x^2-15)) = A/x + B/x^2 + (cx+d) / (x^2-15)
anonymous
  • anonymous
interestingly enough, A = 0
anonymous
  • anonymous
so its equal to integra (-1/(15x) + (1/(15(x^2-15))
anonymous
  • anonymous
x^2=t substitution would save time. By the way, i hope the integrations are clear... i hav to leave...
anonymous
  • anonymous
but you have to do a little work with the substitution
anonymous
  • anonymous
yes..either way its correct.
anonymous
  • anonymous
oh i think you got it
anonymous
  • anonymous
so you fixed the substitution?
anonymous
  • anonymous
yes..
anonymous
  • anonymous
but thats weird
anonymous
  • anonymous
ok x^2 = y, so dx = dy/(2x) = dy/(2sqrt y )
anonymous
  • anonymous
so int 1/(x^2(x^2-15)) = int (1/( 2 sqrt y *y *(y-15))
anonymous
  • anonymous
i used the substitution only for the partial fractions. not for the integrations. i am working with y in partial fracs but wreturning to x in integration
anonymous
  • anonymous
oh
anonymous
  • anonymous
it saves time in partial fracs..
anonymous
  • anonymous
so you got integral 1/(x^2(x^2-15)) = integral (-1/(15x) + (1/(15(x^2-15)). and THEN did substitution
anonymous
  • anonymous
no ..i used y=x^2 only for partial fracs..
anonymous
  • anonymous
you cant do that, as far as i know
anonymous
  • anonymous
That is allowed.
anonymous
  • anonymous
oh then you substituted back before you integrated?
anonymous
  • anonymous
working with a smaller degree variable in partial frac.. is very good idea. yah i subst. back before i integrated
anonymous
  • anonymous
ok so you got integral 1/(x^2(x^2-15)) = integral (-1/(15y) + (1/(15(y-15)).
anonymous
  • anonymous
Yes. that's right. i then changed y to x^2
anonymous
  • anonymous
thats the same thing as just integral 1/(x^2(x^2-15)) = integral (-1/(15x^2) + (1/(15(x^2-15))
anonymous
  • anonymous
oh i see, since its similiar to 1 / ( x (x-5))
anonymous
  • anonymous
yes...but during partial fracs.. dealing with y is easier than dealing with x^2...its just the smae thing
anonymous
  • anonymous
you are right
anonymous
  • anonymous
right, but one has to be careful to switch back
anonymous
  • anonymous
yes...i did a silly mistake back 15 mins. earlier
anonymous
  • anonymous
youre basically using the identity 1/(u(u-15) = -1/15u + 1/15(u-15)
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok
anonymous
  • anonymous
substitution is way for complicated, i mean u substitution
anonymous
  • anonymous
yes, but helps sometimes...avoid it when unnecessary...
anonymous
  • anonymous
right, if we did u substitution, we would be able to do partial fractions anymore
anonymous
  • anonymous
we would have u = x^2 , du = 2x dx, and dx = du/(2x) = du / (2 sqrt u) int ( 1/ (sqrt u * u * (u-15)) du
anonymous
  • anonymous
yes..
anonymous
  • anonymous
this is a very interesting problem, the pitfalls
anonymous
  • anonymous
pitfall is like quicksand, lol
anonymous
  • anonymous
so its better to say, i used the algebraic identity, blah blah
anonymous
  • anonymous
one has to be lucky to get the right idea for integration
anonymous
  • anonymous
so you dont confuse , lol
anonymous
  • anonymous
i substituted into the algebraic identity, that would be best :)
anonymous
  • anonymous
1/(u*(u-15)) = 1/ (15 u ) + ./.. etc
anonymous
  • anonymous
whatever...everyone has a different way to approach problems ..:)
anonymous
  • anonymous
thats fine
anonymous
  • anonymous
its just when you say substitution and integration, most people would think you are doing u substitution
anonymous
  • anonymous
really what it should be called is change of variable
anonymous
  • anonymous
anyways, i see now why partial fractions fail on the complete u substitution
anonymous
  • anonymous
its not a rational function, as you pointed out. thanks for your help. you've been great!
anonymous
  • anonymous
yah...i should have shown the whole process but takes too much time in latex..then there would have been no confusions...by the way i m not too good in english
anonymous
  • anonymous
english aint bad , lol
anonymous
  • anonymous
what country?
anonymous
  • anonymous
where are u from? i am from india
anonymous
  • anonymous
USA
anonymous
  • anonymous
near new york
anonymous
  • anonymous
k...celebrations there right? after osama was killed?
anonymous
  • anonymous
i didnt celebrate
anonymous
  • anonymous
he wasnt given a trial, that kind of thing
anonymous
  • anonymous
i would like to see the evidence, and charges. but i guess it was stupid for him to take credit for it, if he really did take credit
anonymous
  • anonymous
he could've escaped if he was allowed to live....nobody can predict him and his associates,....:)
anonymous
  • anonymous
but he was a CIA asset, when he was fighting the russians
anonymous
  • anonymous
im not crazy about our govt. too much deception
anonymous
  • anonymous
i dont know if he is solely responsible for 911 , he might just be one piece of the puzzle
anonymous
  • anonymous
whatever lets leave these to the concerned authorities... Hope u r ok with par fracs...bye
anonymous
  • anonymous
take care

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