Question

Partial fractions...?

Answer 1

decomposition?

Answer 2

yes, undoing a fraction into its baser parts

Answer 3

\[\int\limits\limits \frac{1}{16x^2 -1} dx\]

Answer 4

16x^2 - 1 is a diff of squares

Answer 5

(4x+1)(4x-1)

Answer 6

Well I'm konfused how exactly they went from the x^2-5x+6 to (x-5/2^2) and (1/2)^2?

Answer 7

looks like they completed a square to get that

Answer 8

oh hmm...

Answer 9

x^2 -5x +6

x^2 -5x +5/2^2 -5/2^2 + 6

(x - 5/2)^2 -5/2^2 + 6

(x - 5/2)^2 - 1/4

Answer 10

wuh haha.,

Answer 11

that 2nd line...

Answer 12

5/4 - 5/4 + 6?

Answer 13

completing the square comes down to adding and subtracting half the square of the "x" coeff

Answer 14

what's the formula for that? haha.

Answer 15

x^2 + bx + c ; to complete the square, add zero to it b/2^2 - b/2^2

Answer 16

oh hmmm.

Answer 17

spose you take a general binomial and square it:

(x+b)^2 = (x+b)(x+b)
= x^2 + 2bx + b^2

notice that the end is have the middle, and squared

Answer 18

(2b/2)^2 = b^2

Answer 19

makes sense.

Answer 20

so, in order to complete a square; take half of the middle and square it :)

but we dont want to change the original value of the problem, so we essentially add zero to the whole thing; but in a useful fashion

b/2^2 - b/2^2 = 0

Answer 21

what you showed above wasn't the middle, but the end?

Answer 22

the b^2?

Answer 23

\[x^2 -5x +6\]
\[x^2 -5x+(0) +6\]
\[x^2 -5x+[(\frac{5}{2})^2-(\frac{5}{2})^2] +6\]
\[[x^2 -5x+(\frac{5}{2})^2]+[-(\frac{5}{2})^2 +6]\]

Answer 24

im bad at keeping my variables in order :)

Answer 25

by convention, if we start out with x^2 + bx + c .... we add and subtract (b/2)^2 was what I was thinking about

Answer 26

it all makes sense tho :P

Answer 27

cool :)

Answer 28

now onto the evil problem :P

Answer 29

so now it's

\[\int\limits \frac{1}{(4x+1)(4x-1)} dx\]

Answer 30

recall that when adding fractions we like to get a common denominator

Answer 31

\[\frac{1}{a}+\frac{1}{b}=\frac{b+a}{ab}\]

Answer 32

:O?

Answer 33

we have the ab part already, so we need to split it up

Answer 34

b+a/ab huh....

Answer 35

\[\frac{1}{(4x+1)(4x-1)}=\frac{A}{4x+1}+\frac{B}{4x-1}\]
this is the usual way of decomposing a fraction

Answer 36

now the methods diverge; i tend to like the cover-up method

Answer 37

but, you could also go the usual method and try to add them back together now

Answer 38

Oh this crap.. I hate it :(

Answer 39

This is where we find a value for A and B right? with using a value for x?

Answer 40

yes

Answer 41

the cover up goes, zero one of the denoms out

Answer 42

x = 1/4?

Answer 43

and -1/4?

Answer 44

yes, and also -1/4

Answer 45

but isnt' a 0 in the denom undefined?

Answer 46

yes, so we ignore that one in the process

Answer 47

oh ok..

Answer 48

the one we ignore is the one we are determining

Answer 49

lets say x=1/4

Answer 50

okies

Answer 51

\[\frac{1}{(4(\frac{1}{4})+1)\cancel{(4(\frac{1}{4}-1))}}=\cancel{\frac{A}{4x+1}}+\frac{B}{(...)}\]

Answer 52

the "resolve" method is more intuitive i think

Answer 53

not sure i learned it like that..

Answer 54

but maybe i don't really recall much except that we want to single out A B C and D....

Answer 55

set everything else to 0 and try to find or something like that...

Answer 56

\[\frac{A}{4x+1}+\frac{B}{4x-1}=\frac{1}{(4x+1)(4x-1)}\]
\[\frac{A(4x-1)}{(4x+1)(4x-1)}+\frac{B(4x+1)}{(4x+1)(4x-1)}=\frac{1}{(4x+1)(4x-1)}\]
\[\frac{A(4x-1)+B(4x+1)}{(4x+1)(4x-1)}=\frac{1}{(4x+1)(4x-1)}\]

all things being equal, compare the tops

Answer 57

oh yeah I think that's how we did it...

Answer 58

4Ax - A + 4Bx + 1 = 1

4(A+B)x + (B-A) = 0x + 1

Answer 59

B-A = 1
B+A = 0

by comparing parts

Answer 60

B-A = 1
B+A = 0
--------
2B = 1

B = 1/2

therefore A = -1/2

Answer 61

where does that b-a come from? I see the extra 'A' but what abotu that B?

Answer 62

in the last step of the fancy stuff, expand out the numerator

A(4x-1) + B(4x-1) = 1 is what we end up with

Answer 63

err, one of those is a +1 :)

4Ax - A + 4Bx + B = 1

Answer 64

now we want to gather together our x parts and our constant parts to compare with

Answer 65

4Ax + 4Bx = 4(A+B)x

-A + B = (B-A)

Answer 66

compare these to the other side of the equals sign; where we have 0x + 1

Answer 67

4(A+B)x = 0x when A+B = 0

(B-A) = 1 well, when B-A = 1

Answer 68

I gotta read up h/o haha.

Answer 69

im outta MtDew :)

Answer 70

I feel like this is super extra complicated...

Answer 71

in my book all we had to do was set x = to something, and then solve 1 at a time....

Answer 72

if you never learned the proper way to add fractions, then yes; it does seem a bit mixed up

Answer 73

oh, your book gives the cross out method; works fine too

Answer 74

there comes a point when the cross out fails tho, so youd have to finish up with this method

Answer 75

maybe

Answer 76

It just seems complicated, prob cuz my head wants to explode LOL

:(

Answer 77

that page is the method we just did

Answer 78

*facepalm*

Answer 79

just think of it as adding fractions; cause thats all it amounts to

Answer 80

okies

Answer 81

find a common denominator ... that parts actually already done; you just got to remember how to finish the adding

Answer 82

After this problem I tihkn I'll leave the rest for a time when my brain doesn't want to kerplode :P

Answer 83

usually tames down after a half gallon of happy tracks icecream

Answer 84

:)

Answer 85

so A+B =0 and B-A = 1

Answer 86

in the end, yes

Answer 87

now its just a system of 2 eqs in 2 unknowns

Answer 88

hmm

Answer 89

Idk where we even are in this lol.

Answer 90

so why exactly did we solve for A+B and B-A?

Answer 91

becasue those are our missing numerators when we split it into 2 fractions

Answer 92

finding A and B amounts to determining the splited fractions that equate to the original whole thing

Answer 93

oh okay we got those bad boys before :P.

Answer 94

when we split it up I see...

Answer 95

yes, the whole thing amounts to finding the numerators that fit the split up

Answer 96

there are also finer points in regards to how it can split as well ... but fer now, this is fine