Question

question about setting up a partial fraction for integration.

Answer 1

In class, I was told that \[\int\limits_{indef} x^2+3/(x^3+2x)\] could be simplified by using \[x^2+3=(A/x)+(Bx+C)/(x^2+2)\]

How do I know when to use A vs using BX+C?

Answer 2

is there a rule or something? in the examples, I've seen a/x, or bx+c/x^2 or whatever the reduced denominator is.

Answer 3

when there is a function of degree 1 in denominator, use constant A (or something..)
when there is a function of degree 2 in denominator, use linear function Bx+C.

Answer 4

only when that degree 2 polynomial cannot be factorised further

Answer 5

ah ok... Does it have to be a degree 2 polynomial? What if you had something with x^3?

Answer 6

as i said, factorize as much as possible.
if u cannot factorize that 3rd degree polynomial, x^3...
then u can use Ex^2+Fx+G in numerator.

Answer 7

use a polynomial of 1 less degree in numerator

Answer 8

be careful not to drop the denominator from the left prematurely
\[\frac{x ^{2}+3}{x(x ^{2}+2)}=\frac{A}{x}+\frac{Bx+C}{x ^{2}+2}\]

Answer 9

Ah ok I understand now :). Thank you both. So once I factor it as much as possible, I separate and make the denominator one degree lower and then solve for A,B,C

Answer 10

or, not denomoniator.... sorry... typo.... numerator.

Answer 11

Dx.... well... you get what I mean xD

Answer 12

generally, yes

Answer 13

yup.

Answer 14

Thank you both.... I wish I could have chosen both of your responses, but nonetheless, thank you for clearing that up for me :)

Answer 15

i am sorry, i just realized this :
if the polynomial is of degree 3, then it must have one real root(say k)
so, u must be able to factor out (x-k) and u get one linear and one quadratic polynomial .....
if the polynomial is of degree 4, then you must be able to factor it as 2 quadratic polynomials(if it has 4 complex roots) and you get 2 quadratic polynomials.
and so on for higher degree....
so u will NEVER get a situation with degree 3 or more polynomial in denominator, which cannot be factorized....
so u will never use quadratic in numerator(only A,Ax+B)
hope u understand this.

Answer 16

Is it possible to get a fourth degree polynomial that has only complex roots and therefore does not have to quadratic factors?

Answer 17

*two... quadratic factors?

Answer 18

I guess not... just foiled in my head.

Answer 19

if it has 2 complex roots then it can be factored as (Ax^2+Bx+C)(Dx+E)(Fx+G)
because then there will be 2 real roots(-E/D and -G/F here)

Answer 20

conjugate pairs should always result in a quadratic if not a constant.... oh, the wonderful complexities.

Answer 21

yup, 'conjugate pairs should always result in a quadratic' \(\checkmark\)

Answer 22

I would become a fan again if I could ;)

Answer 23

ah ok.... I think I understand.... and if not, I'm not at much of a loss... I know how to do it, so that's all that matters for now xD.