## sjg13e one year ago Partial Functions - Help!

1. sjg13e

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

2. anonymous

it's alright

3. anonymous

i wait ....

4. sjg13e

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

5. sjg13e

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

6. amistre64

apart from a typo or two its going fine so far

7. sjg13e

then $x^2 - 3 = A(x+2) = B(x-2)$ $x^2 - 3 = Ax + 2A + Bx - 2B$ $x^2 - 3 = (A + B)x - 2A - 2B$

8. sjg13e

so yes, sorry there are some typos!

9. amistre64

dont bother distributing .. let x=2 and -2 to solve for A and B $(2)^2 - 3 = A(2+2) + B(2-2)$ $(-2)^2 - 3 = A(-2+2) + B(-2-2)$

10. sjg13e

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 (??)

11. freckles

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

12. amistre64

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

13. sjg13e

@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

14. amistre64

well, you saw why distribution cuases issues already :)

15. amistre64

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

16. amistre64

(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

17. freckles

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

18. freckles

19. amistre64

x^2-3 + (1-1) = (x^2-4) + 1 if you really want to do the division first, but its not required

20. sjg13e

@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

21. amistre64

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.

22. sjg13e

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

23. amistre64

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

24. Jhannybean

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

25. amistre64

Jhanny, its a conspiracy :)

26. Jhannybean

Knew it!

27. amistre64

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

28. sjg13e

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

29. freckles

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

30. sjg13e

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