Question

Partial Functions - Help!

Answer 1

please give me a second to upload the question and my work :)

Answer 2

it's alright

Answer 3

i wait ....

Answer 4

\[\int\limits_{}^{}\frac{ x^2 - 3 }{ x^2 - 4 }\]

Answer 5

my work (it's wrong obviously lol):
\[(x^2 - 4) = (x - 2)(x +2)\]
so
\[\int\limits_{}^{}\frac{ x^2 - 3 }{ x^2 - 4 }= \frac{ x^2 - 3 }{ (x^2 - 2)(x^2 +2) } = \frac{ A }{ x -2 }+\frac{ B }{ x + 2 }\]

Answer 6

apart from a typo or two its going fine so far

Answer 7

then
\[x^2 - 3 = A(x+2) = B(x-2) \]
\[x^2 - 3 = Ax + 2A + Bx - 2B\]
\[x^2 - 3 = (A + B)x - 2A - 2B\]

Answer 8

so yes, sorry there are some typos!

Answer 9

dont bother distributing .. let x=2 and -2 to solve for A and B

\[(2)^2 - 3 = A(2+2) + B(2-2)\]
\[(-2)^2 - 3 = A(-2+2) + B(-2-2)\]

Answer 10

From there, I tried to find the values of A and B
A + B = 0
A = -B
\[-2A - 2B = -3 \]
\[-2A = -3 + 2(-A)\]
\[-2A = -3 - 2A\]
0 = 3 (??)

Answer 11

Partial most likely won't work until you do division. We need to divide first because you have an improper fraction.

Answer 12

... it works fine, just dont do the distribution part; that way we can zero out a term to solve

Answer 13

@amistre64 Wait, can you explain a little more why I shouldn't bother distributing and should instead just use substitution.
@freckles, How did you know that I need to divide first? That step didn't immediately pop out to me

Answer 14

well, you saw why distribution cuases issues already :)

Answer 15

A(x-2) can be eliminated when x=2

B(x+2) can be eliminated when x=-2

Answer 16

(2)^2-3=A(2+2)+B(2-2)

4-3=A(4)+B(0)

1 = 4A
.........................

(-2)^2-3=A(-2+2)+B(-2-2)

4-3=A(0)+B(-4)

1=-4B

Answer 17

Your top poly and bottom poly had equal degree. Whenever the top poly has equal or greater degree to that of the bottom poly. I think division then partial

Answer 18

Anyways i will let Amistre help you because it is hard to read on this phone

Answer 19

x^2-3 + (1-1) = (x^2-4) + 1

if you really want to do the division first, but its not required

Answer 20

@amistre64 okay, i understand the following work, but how did you know to let x = 2? in my calculus class we haven't really discussed this method within partial functions

and thank you @freckles for your help! i'll try doing division first

Answer 21

well, how else do you zero out a term?

A(x+2) + B(x-2)

anything times 0 is 0 ... so by observation letting x=2 or -2 gets rid of a term so that we can solve for the other.

Answer 22

okay, that makes sense! thank you! i'll try that method and solve the problem again

Answer 23

\[\frac{x^2-3}{x^2-4}\implies \frac{(x^2-4)+1}{x^2-4}\implies1+\frac{1}{x^2-4}\]
we still have to work the partial fraction even after the division, so the division is not necessary to me.

Answer 24

What i realized for most functions with polynomials in the denominator, it's either u-sub or partial fraction decomp. Hmm!

Answer 25

Jhanny, its a conspiracy :)

Answer 26

Knew it!

Answer 27

hmm, working it to a proper fraction might be needed ... never considered it before but the wolf suggests that it might be needed.

Answer 28

okay, so i would have to do long division before i decompose it?

Answer 29

yes @sjg13e
as we notice
\[\frac{A}{x-2}+\frac{B}{x+2}=\frac{(A+B)x+(2A-2B)}{x^2-4} \neq \frac{x^2-3}{x^2-4} \text{ for any real choice } \\ A \text{ and } B\]
so division is definitely required before doing the partial fractions

Answer 30

Thank you @freckles! I solved the problem and read through some calculus notes and realized that it's required to get the leading x degree in the numerator to be less than the leading x degree in the denominator, and for this problem, it required long division