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anonymous
 5 years ago
integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^215))
should i do partial fraction
1/((x^2*(x^215)) = A / x + B/x^2 + C/(xsqrt 15) + D/(x+sqrt 15) , or
1/((x^2*(x^215)) = A / x + B/x^2 + Cx+d / ( x^2  15)
anonymous
 5 years ago
integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^215)) should i do partial fraction 1/((x^2*(x^215)) = A / x + B/x^2 + C/(xsqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^215)) = A / x + B/x^2 + Cx+d / ( x^2  15)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why don't you type in math. will help in viewing easily.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0whats wrong with partial fractions with radicals?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont see any problem with the first way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problems but we avoid it as calculations get messy.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my computer does the partial fractions integral 1/(x^2 (x^215)) = 1/(15x^2) + 1/(15(x^215))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah...that "seems" easy :) you can always take x^2=y during partial fracs.. then u get 1/(y(y15)) which can be easily split into fracs.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that doesnt disobey partial fractions rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought it was , it had to be linear factors and irreducibles

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like A/x + B/x^2 + C / (xsqrt 15) + D/(x+sqrt 15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0See...there are no socalled rules...at least not for me :) if we take radicals the math is correct but its unconventional

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cx+d is supposed to be only for irreducibles, i thought

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like integral 1/x(x^2+1) = A/x + bx+c / x^2 + 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so youre saying 1/x(x^21) = A/x + (bx+c) /( x^2  1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not only for irreducibles,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im pretty sure thats what the theorem states

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe the theorem u saw said that we cx+d for irreducibles but it did not say that we cant use cx+d for nonirreducibles!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we use cx+d for irreducibles but we can also use it like (bx+c)/(x^21). Here the math is correct. But splitting like A/(x+1)+B/(x1) is more advantageous, as it gives a good form whose integral is known.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for example, this partial fractions fails 1/(x^2 (x5) ) = A/x^2 + B/(x5) , this fails

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh because of the A/x + B/x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so 1/[(xa)^n (x^2 +bx +c)^n] = k1/(xa) + k2/(xa)^2 + ... kn/(xa)^n + m1x+p1/(x^2 +bx +c) +...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your upto kn is ok. but from m1,p1,...its getting complicated.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah im running out of letters

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no. there will be cx+d term above and the denominator's power will increase from 1 to n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/[(xa)^n (x^2 +bx +c)^n] = A1/(xa) + A2/(xa)^2 + ... An/(xa)^n + B1x+C1/(x^2 +bx +c) + (B2x +C2)/(x^2 +bx + c)^2... +

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/[(xa)^n (x^2 +bx +c)^n] = A1/(xa) + A2/(xa)^2 + ... An/(xa)^n + B1x+C1/(x^2 +bx +c) + (B2x +C2)/(x^2 +bx + c)^2... +(Bnx +Cn) / (x^2 +bx +c)^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your c1 and b1 terms are wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there won't be any b1x ... also instead of c1 write it as b1x+c1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/[(xa)^n (x^2 +bx +c)^n] = A1/(xa) + A2/(xa)^2 + ... An/(xa)^n + (B1x+C1)/(x^2 +bx +c) + (B2x +C2)/(x^2 +bx + c)^2... +(Bnx +Cn) / (x^2 +bx +c)^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yups. thats correct....:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i left out the parenthesees

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so thats fine for partial fractions , and x^2 +bx + c doesnt necessarily have to be irreducible,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see..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
 5 years ago
Best ResponseYou've already chosen the best response.0a reducible quadratic you mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0reducible into linear factors

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yups yups a QUADRATIC can only have cx+d above it. yes reducible into linear REAL factors. Nobody wants to have complex factors here!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so then we could do 1/(x^2 (x^2  15)) , we could reduce it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im 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
 5 years ago
Best ResponseYou've already chosen the best response.0Mathematically permissible. But not recommended.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets try this one 1/ ( x (x^2+2x+1)) integrate that using parfrac

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0int 1/ ( x (x^2+2x+1))= (AX+b) / ( x^2 + 2x + 1) + C / x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in part fracs.. its (1/x)+((x2)/(x^2+2x+1)) the first term gives logx and the second make substitution x^2+2x=y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, or we could have factored x^2 + 2x + 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats that substitution ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yupss....the subs wrong...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh by the way, partial fraction doesnt help us with say 1/ (x^(3/2) (x15))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no.. since the denominator is not a POLYNOMIAL

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0both numer. and denom. must be polynomials

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait is that a precondition?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you could have 1/(x*(x^2sqrt 3)) for example

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YES partial fractions can help ONLY TO RATIONAL FUNCTIONS i.e. both numeratoer and denominator polynomials with denominator not being the zero polynomial

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just clearing that up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0by the way what's your age?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hahha..joking? i hope u understood.. ask for doubts...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0jeez, i thought you were older. you are wise beyond your years :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you a mathematician? or mathrelated worker? u named urself cantorset...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for a living, i was stumped because the student didnt want to reduce a quadratic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and i thought, hey the theorem says so and so

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it was reducible into irrational factors

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the question u asked me?  yes it can be solved both ways

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0see this problem we just did its a lot easier if we reduce the quadratic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral (1/(x(x^2+2x+1))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes 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
 5 years ago
Best ResponseYou've already chosen the best response.0right so i guess it depends on the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well its good to know the power of this theorem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which area of math are you specialized in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i tutor mostly calculus ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow..thats awesome..i love calculus..but know little..struggling still with single variable calculus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well if you need help, give me a buzz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so yeah, this one is tougher to integrate , without reducing the quadratic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integral (1/x)+((x2)/(x^2+2x+1))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the original question integral 1/(x^2 (x^215))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then reducing the quadratic is worse, lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2=y, integr=(1/15)log(y15)/y+constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you need 2x dx = dy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh its not so simple ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i.e. integral=\[\frac{1}{15}\ln \left \frac{x^215}{x^2} \right+constant\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you make a substitution, you have to do dy = 2x dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0SORRY!! out of my mind......u r right..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then we have another problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we have dx = dy / (2sqrt y))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, lets not do the substitution then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{15x}+\frac{1}{30\sqrt{15}}\ln \left \frac{x\sqrt{15}}{x+\sqrt{15}} \right+constant\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think this is the answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/(15*x)(1/225)*sqrt(15)*arctanh((1/15)*x*sqrt(15))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what did you use ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}\frac{dx}{x^215}=\frac{1}{\sqrt{15}} \ln \left \frac{x\sqrt{15}}{x+\sqrt{15}} \right +C\] i used this identity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but theres an x^2 in front

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{?}^{?}1/(x^2(x^215))\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you do fraction in latex?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/(x^2(x^215)) = A/x + B/x^2 + (cx+d) / (x^215)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0interestingly enough, A = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its equal to integra (1/(15x) + (1/(15(x^215))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2=t substitution would save time. By the way, i hope the integrations are clear... i hav to leave...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but you have to do a little work with the substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes..either way its correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i think you got it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you fixed the substitution?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok x^2 = y, so dx = dy/(2x) = dy/(2sqrt y )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so int 1/(x^2(x^215)) = int (1/( 2 sqrt y *y *(y15))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0it saves time in partial fracs..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you got integral 1/(x^2(x^215)) = integral (1/(15x) + (1/(15(x^215)). and THEN did substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no ..i used y=x^2 only for partial fracs..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you cant do that, as far as i know

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh then you substituted back before you integrated?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0working with a smaller degree variable in partial frac.. is very good idea. yah i subst. back before i integrated

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so you got integral 1/(x^2(x^215)) = integral (1/(15y) + (1/(15(y15)).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. that's right. i then changed y to x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats the same thing as just integral 1/(x^2(x^215)) = integral (1/(15x^2) + (1/(15(x^215))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see, since its similiar to 1 / ( x (x5))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes...but during partial fracs.. dealing with y is easier than dealing with x^2...its just the smae thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, but one has to be careful to switch back

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes...i did a silly mistake back 15 mins. earlier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre basically using the identity 1/(u(u15) = 1/15u + 1/15(u15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0substitution is way for complicated, i mean u substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but helps sometimes...avoid it when unnecessary...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, if we did u substitution, we would be able to do partial fractions anymore

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we would have u = x^2 , du = 2x dx, and dx = du/(2x) = du / (2 sqrt u) int ( 1/ (sqrt u * u * (u15)) du

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is a very interesting problem, the pitfalls

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0pitfall is like quicksand, lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its better to say, i used the algebraic identity, blah blah

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one has to be lucky to get the right idea for integration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you dont confuse , lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i substituted into the algebraic identity, that would be best :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/(u*(u15)) = 1/ (15 u ) + ./.. etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whatever...everyone has a different way to approach problems ..:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its just when you say substitution and integration, most people would think you are doing u substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0really what it should be called is change of variable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anyways, i see now why partial fractions fail on the complete u substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its not a rational function, as you pointed out. thanks for your help. you've been great!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah...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
 5 years ago
Best ResponseYou've already chosen the best response.0english aint bad , lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where are u from? i am from india

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k...celebrations there right? after osama was killed?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he wasnt given a trial, that kind of thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0he could've escaped if he was allowed to live....nobody can predict him and his associates,....:)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but he was a CIA asset, when he was fighting the russians

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not crazy about our govt. too much deception

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know if he is solely responsible for 911 , he might just be one piece of the puzzle

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whatever lets leave these to the concerned authorities... Hope u r ok with par fracs...bye
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