Question

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)

Answer 1

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

Answer 2

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!!!)

Answer 3

hello

Answer 4

whats wrong with partial fractions with radicals?

Answer 5

i dont see any problem with the first way

Answer 6

no problems but we avoid it as calculations get messy.

Answer 7

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

Answer 8

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.

Answer 9

that doesnt disobey partial fractions rule

Answer 10

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

Answer 11

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

Answer 12

See...there are no so-called rules...at least not for me :) if we take radicals the math is correct but its unconventional

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

not only for irreducibles,

Answer 17

im pretty sure thats what the theorem states

Answer 18

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!!

Answer 19

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.

Answer 20

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

Answer 21

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

Answer 22

yes

Answer 23

hmmm

Answer 24

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) +...

Answer 25

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

Answer 26

yeah im running out of letters

Answer 27

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

Answer 28

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... +

Answer 29

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

Answer 30

your c1 and b1 terms are wrong

Answer 31

why?

Answer 32

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

Answer 33

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

Answer 34

yups. thats correct....:)

Answer 35

so thats fine?

Answer 36

oh i left out the parenthesees

Answer 37

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

Answer 38

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?

Answer 39

a reducible quadratic you mean

Answer 40

reducible into linear factors

Answer 41

yups yups a QUADRATIC can only have cx+d above it.
yes reducible into linear REAL factors. Nobody wants to have complex factors here!

Answer 42

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

Answer 43

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

Answer 44

Mathematically permissible. But not recommended.

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

yes...

Answer 50

wait thats wrong

Answer 51

whats that substitution ?

Answer 52

yupss....the subs wrong...

Answer 53

we can do it however

Answer 54

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

Answer 55

no.. since the denominator is not a POLYNOMIAL

Answer 56

, ahh

Answer 57

both numer. and denom. must be polynomials

Answer 58

wait is that a precondition?

Answer 59

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

Answer 60

YES partial fractions can help ONLY TO RATIONAL FUNCTIONS i.e. both numeratoer and denominator polynomials with denominator not being the zero polynomial

Answer 61

ok

Answer 62

just clearing that up

Answer 63

by the way what's your age?

Answer 64

25

Answer 65

why do i sound dumb

Answer 66

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

Answer 67

i am 17

Answer 68

oh

Answer 69

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

Answer 70

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

Answer 71

yes im a math tutor

Answer 72

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

Answer 73

and i thought, hey the theorem says so and so

Answer 74

it was reducible into irrational factors

Answer 75

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

Answer 76

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

Answer 77

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

Answer 78

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..

Answer 79

right
so i guess it depends on the problem

Answer 80

well its good to know the power of this theorem

Answer 81

which area of math are you specialized in?

Answer 82

i tutor mostly calculus ,

Answer 83

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

Answer 84

youre amazing :)

Answer 85

well if you need help, give me a buzz

Answer 86

ok..thank you.

Answer 87

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

Answer 88

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

Answer 89

yups.

Answer 90

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

Answer 91

then reducing the quadratic is worse, lol

Answer 92

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

Answer 93

then you need
2x dx = dy

Answer 94

oh its not so simple ,

Answer 95

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

Answer 96

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

Answer 97

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

Answer 98

then we have another problem