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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- anonymous

why don't you type in math. will help in viewing easily.

- 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

hello

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## More answers

- anonymous

whats wrong with partial fractions with radicals?

- anonymous

i dont see any problem with the first way

- anonymous

no problems but we avoid it as calculations get messy.

- anonymous

my computer does the partial fractions
integral 1/(x^2 (x^2-15)) = -1/(15x^2) + 1/(15(x^2-15))

- 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

that doesnt disobey partial fractions rule

- anonymous

i thought it was , it had to be linear factors and irreducibles

- anonymous

like A/x + B/x^2 + C / (x-sqrt 15) + D/(x+sqrt 15)

- 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

cx+d is supposed to be only for irreducibles, i thought

- anonymous

like integral 1/x(x^2+1) = A/x + bx+c / x^2 + 1

- anonymous

so youre saying
1/x(x^2-1) = A/x + (bx+c) /( x^2 - 1)

- anonymous

not only for irreducibles,

- anonymous

im pretty sure thats what the theorem states

- 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

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

for example, this partial fractions fails
1/(x^2 (x-5) ) = A/x^2 + B/(x-5) , this fails

- anonymous

oh because of the A/x + B/x^2

- anonymous

yes

- anonymous

hmmm

- 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

your upto kn is ok. but from m1,p1,...its getting complicated.

- anonymous

yeah im running out of letters

- anonymous

no. there will be cx+d term above and the denominator's power will increase from 1 to n

- 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

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

your c1 and b1 terms are wrong

- anonymous

why?

- anonymous

there won't be any b1x ...
also instead of c1 write it as b1x+c1

- 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

yups. thats correct....:)

- anonymous

so thats fine?

- anonymous

oh i left out the parenthesees

- anonymous

ok so thats fine for partial fractions , and x^2 +bx + c doesnt necessarily have to be irreducible,

- 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

a reducible quadratic you mean

- anonymous

reducible into linear factors

- 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

ok, so then we could do 1/(x^2 (x^2 - 15)) , we could reduce it

- 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

Mathematically permissible. But not recommended.

- anonymous

ok lets try this one
1/ ( x (x^2+2x+1)) integrate that using parfrac

- anonymous

int 1/ ( x (x^2+2x+1))= (AX+b) / ( x^2 + 2x + 1) + C / x

- 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

right, or we could have factored x^2 + 2x + 1

- anonymous

yes...

- anonymous

wait thats wrong

- anonymous

whats that substitution ?

- anonymous

yupss....the subs wrong...

- anonymous

we can do it however

- anonymous

oh by the way, partial fraction doesnt help us with say 1/ (x^(3/2) (x-15))

- anonymous

no.. since the denominator is not a POLYNOMIAL

- anonymous

, ahh

- anonymous

both numer. and denom. must be polynomials

- anonymous

wait is that a precondition?

- anonymous

you could have 1/(x*(x^2-sqrt 3)) for example

- 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

ok

- anonymous

just clearing that up

- anonymous

by the way what's your age?

- anonymous

25

- anonymous

why do i sound dumb

- anonymous

hahha..joking? i hope u understood.. ask for doubts...

- anonymous

i am 17

- anonymous

oh

- anonymous

jeez, i thought you were older. you are wise beyond your years :)

- anonymous

are you a mathematician? or math-related worker? u named urself cantorset...

- anonymous

yes im a math tutor

- anonymous

for a living, i was stumped because the student didnt want to reduce a quadratic

- anonymous

and i thought, hey the theorem says so and so

- anonymous

it was reducible into irrational factors

- anonymous

the question u asked me? - yes it can be solved both ways

- anonymous

see this problem we just did
its a lot easier if we reduce the quadratic

- anonymous

integral (1/(x(x^2+2x+1))

- 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

right
so i guess it depends on the problem

- anonymous

well its good to know the power of this theorem

- anonymous

which area of math are you specialized in?

- anonymous

i tutor mostly calculus ,

- anonymous

wow..thats awesome..i love calculus..but know little..struggling still with single variable calculus

- anonymous

youre amazing :)

- anonymous

well if you need help, give me a buzz

- anonymous

ok..thank you.

- anonymous

so yeah, this one is tougher to integrate , without reducing the quadratic

- anonymous

integral (1/x)+((-x-2)/(x^2+2x+1))

- anonymous

yups.

- anonymous

so the original question
integral 1/(x^2 (x^2-15))

- anonymous

then reducing the quadratic is worse, lol

- anonymous

x^2=y, integr=(1/15)log|(y-15)/y|+constant

- anonymous

then you need
2x dx = dy

- anonymous

oh its not so simple ,

- anonymous

i.e. integral=\[\frac{1}{15}\ln \left| \frac{x^2-15}{x^2} \right|+constant\]

- anonymous

when you make a substitution, you have to do dy = 2x dx

- anonymous

SORRY!! out of my mind......u r right..

- anonymous

then we have another problem

- anonymous

we want dx by itself

- anonymous

dy/(2x) = dx

- anonymous

so we have dx = dy / (2sqrt y))

- anonymous

ok, lets not do the substitution then

- anonymous

\[\frac{1}{15x}+\frac{1}{30\sqrt{15}}\ln \left| \frac{x-\sqrt{15}}{x+\sqrt{15}} \right|+constant\]

- anonymous

i think this is the answer.

- anonymous

let me check

- anonymous

i use maple

- anonymous

1/(15*x)-(1/225)*sqrt(15)*arctanh((1/15)*x*sqrt(15))

- anonymous

must be the same..

- anonymous

oh i messed up

- anonymous

so what did you use ?

- 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

but theres an x^2 in front

- anonymous

that is easy!!

- anonymous

\[\int\limits_{?}^{?}1/(x^2(x^2-15))\]

- anonymous

how do you do fraction in latex?

- anonymous

frac{2}{3}=2/3

- anonymous

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

- anonymous

interestingly enough, A = 0

- anonymous

so its equal to integra (-1/(15x) + (1/(15(x^2-15))

- anonymous

x^2=t substitution would save time. By the way, i hope the integrations are clear... i hav to leave...

- anonymous

but you have to do a little work with the substitution

- anonymous

yes..either way its correct.

- anonymous

oh i think you got it

- anonymous

so you fixed the substitution?

- anonymous

yes..

- anonymous

but thats weird

- anonymous

ok x^2 = y, so dx = dy/(2x) = dy/(2sqrt y )

- anonymous

so int 1/(x^2(x^2-15)) = int (1/( 2 sqrt y *y *(y-15))

- 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

oh

- anonymous

it saves time in partial fracs..

- anonymous

so you got integral 1/(x^2(x^2-15)) = integral (-1/(15x) + (1/(15(x^2-15)). and THEN did substitution

- anonymous

no ..i used y=x^2 only for partial fracs..

- anonymous

you cant do that, as far as i know

- anonymous

That is allowed.

- anonymous

oh then you substituted back before you integrated?

- anonymous

working with a smaller degree variable in partial frac.. is very good idea.
yah i subst. back before i integrated

- anonymous

ok so you got integral 1/(x^2(x^2-15)) = integral (-1/(15y) + (1/(15(y-15)).

- anonymous

Yes. that's right. i then changed y to x^2

- anonymous

thats the same thing as just
integral 1/(x^2(x^2-15)) = integral (-1/(15x^2) + (1/(15(x^2-15))

- anonymous

oh i see, since its similiar to 1 / ( x (x-5))

- anonymous

yes...but during partial fracs.. dealing with y is easier than dealing with x^2...its just the smae thing

- anonymous

you are right

- anonymous

right, but one has to be careful to switch back

- anonymous

yes...i did a silly mistake back 15 mins. earlier

- anonymous

youre basically using the identity 1/(u(u-15) = -1/15u + 1/15(u-15)

- anonymous

yes

- anonymous

ok

- anonymous

substitution is way for complicated, i mean u substitution

- anonymous

yes, but helps sometimes...avoid it when unnecessary...

- anonymous

right, if we did u substitution, we would be able to do partial fractions anymore

- 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

yes..

- anonymous

this is a very interesting problem, the pitfalls

- anonymous

pitfall is like quicksand, lol

- anonymous

so its better to say, i used the algebraic identity, blah blah

- anonymous

one has to be lucky to get the right idea for integration

- anonymous

so you dont confuse , lol

- anonymous

i substituted into the algebraic identity, that would be best :)

- anonymous

1/(u*(u-15)) = 1/ (15 u ) + ./.. etc

- anonymous

whatever...everyone has a different way to approach problems ..:)

- anonymous

thats fine

- anonymous

its just when you say substitution and integration, most people would think you are doing u substitution

- anonymous

really what it should be called is change of variable

- anonymous

anyways, i see now why partial fractions fail on the complete u substitution

- anonymous

its not a rational function, as you pointed out. thanks for your help. you've been great!

- 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

english aint bad , lol

- anonymous

what country?

- anonymous

where are u from?
i am from india

- anonymous

USA

- anonymous

near new york

- anonymous

k...celebrations there right? after osama was killed?

- anonymous

i didnt celebrate

- anonymous

he wasnt given a trial, that kind of thing

- 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

he could've escaped if he was allowed to live....nobody can predict him and his associates,....:)

- anonymous

but he was a CIA asset, when he was fighting the russians

- anonymous

im not crazy about our govt. too much deception

- anonymous

i dont know if he is solely responsible for 911 , he might just be one piece of the puzzle

- anonymous

whatever lets leave these to the concerned authorities... Hope u r ok with par fracs...bye

- anonymous

take care

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