partial fractions

for 1/(1+x^3),

I've got to = A/(x+1) + B/(x^2 -x+1)

But the values of A and B are different depending on x. So they must include an x term? Where do I go from here?

Reply 1

EwanWest

You need to have a polynomial term on top of the fraction which is 1 order lower than the denominator
EDIT: this is good for most, repeated denominator terms are different though

(edited 7 years ago)

Reply 2

Zacken

It should be

\displaystyle \frac{1}{1+x^3} = \frac{1}{(x+1)(x^2 - x + 1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 - x + 1}

If your partal fraction term has polynomial of degree n in the denominator, then the numerator should be of degree n-1. Which is why linear term at the bottom => constant term on top. Quadratic in bottom (like x^2 - x +1) => linear term on top.

Reply 3

TSR George

Aha thank you both! I'll give it another go :biggrin:

Reply 4

TSR George

is this still the case if the denominator is (ax+b)^2 ?

Reply 5

Zacken

No, if the denominator is

(ax+b)^2

then you do

\frac{1}{(ax+b)^2} = \frac{A}{ax+b} + \frac{B}{(ax+b)^2}

.

http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx - read this.

Reply 6

TSR George

Original post by Zacken

No, if the denominator is

(ax+b)^2

then you do

\frac{1}{(ax+b)^2} = \frac{A}{ax+b} + \frac{B}{(ax+b)^2}

.

http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx - read this.

Thank you

Reply 7

EwanWest

It becomes more complicated. If I had 1/(ax+b)(cx+d)^2 this would be split into A/(ax+b) + B/(cx+d) + (Cx+D)/(cx+d)^2

Reply 8

TSR George

Original post by EwanWest

It becomes more complicated. If I had 1/(ax+b)(cx+d)^2 this would be split into A/(ax+b) + B/(cx+d) + (Cx+D)/(cx+d)^2

But @Zacken just said this would be C/(cx+d)^2 ??

Reply 9

EwanWest

Yes he's correct, I haven't done these in a while. Having said that if you have the extra term in there like I do it will just come out as 0 when you compare coefficients so it won't give you the wrong answer