Question

HELP!!! Integration by Partial Fractions!

1.) integral dx/(x^2+x-2)

2.) integral x^2+x-16/(x+1)(x-3)^2 dx

3.) Integral 2x^2 - 9x - 9/x^3-9x dx

Answer 1

1.) \[\int\limits \frac{ dx }{ x^2+x-2 }\]

2.) \[\int\limits \frac{ x^2+x-16 }{ (x+1)(x-3)^2 } dx\]

3.) \[\int\limits \frac{ 2x^2-9x-9 }{ x^3-9x } dx\]

Answer 2

The partial fractions from Mathematica follow. You do the integrations.\[\left\{\frac{1}{3 (x-1)}-\frac{1}{3 (x+2)},-\frac{1}{x+1}+\frac{2}{x-3}-\frac{1}{(x-3)^2},\frac{1}{x}+\frac{2}{x+3}-\frac{1}{x-3}\right\} \]

Answer 3

So for \(\Large 1\), do you understand how to `factor` the denominator?

Answer 4

Yes I do! It's \[(x+2)(x-1)\]

Answer 5

Ok good good.
The idea is that, there's some process (I wouldn't be able to explain where it comes from :P) that allows us to split up the factors.

The numerators will always be `one degree less than` the denominator.
So in this example, both factors are `linear` (x to the first power), so the unknown numerators should be `constant`

\[\Large \frac{1}{(x+2)(x-1)} \quad=\quad \frac{A}{x+2}+\frac{B}{x-1}\]

Answer 6

Our next step is to multiply both sides by the big chunky denominator on the left side there. k?

Answer 7

Yeah! Sorry for the delay! Open Study with technical problems again! But, so far, so good!

Answer 8

So multiplying both sides by the stuff gives us,\[\Large 1 \quad=\quad A(x-1)+B(x+2)\]
Understand how one of the factors cancelled on the A term and one on the B term?

Answer 9

Yes, I do!

Answer 10

From here there are a couple of different approaches you can take to solve for A and B.
You can distribute everything out, and then equate like terms.

So for this problem:\[\Large 0x+1 \quad=\quad Ax-A+Bx+2B\]So you would equate the x's,\[\Large 0x=Ax+Bx \qquad\to\qquad 0=A+B\]and equate the constants,\[\Large 1=A+2B\]
And from there you have a pretty easy system of equations that allows you to solve for A and B.

I don't particularly care for that method though.
Since both of our factors were linear we can take a nice simple approach.

Answer 11

Let's start by plugging in x=1, that should give us our B value pretty quickly.

Answer 12

\[\Large 1 \quad=\quad A(1-1)+B(1+2)\]See how that gets us our B nice and cleanly? :D

Answer 13

You're trying to use nice values that will zero out one of the variables.

Answer 14

Okay! So, it's B equal to 1/3?

Answer 15

Looks good!
What x value should we use to solve for A?

Answer 16

We can use 1 too, right?

Answer 17

Since we've found our B, we `could` plug in B=1/3, and then reuse the x=1 again, yes.

But a bit of an easier way it to use an x value that will zero out the B term.

Answer 18

Oh, that's true!

Answer 19

\[\Large 1 \quad=\quad A(x-1)+B(x+2)\]

Hmm it looks like x=-2, yes? :o

Answer 20

Yes! x=-2!

Answer 21

\[\Large 1\quad=\quad A(-2-1)+\cancel{B(-2+2)}\]

Answer 22

So, A = -1/3

Answer 23

Cool! c:

Answer 24

So we've determined that:\[\Large \int\limits \frac{1}{(x+2)(x-1)}dx \quad=\quad \int\limits\frac{A}{x+2}+\frac{B}{x-1}dx\]Plugging in our A and B,\[\Large \int\limits \frac{1}{(x+2)(x-1)}dx \quad=\quad \int\limits\frac{-1/3}{x+2}+\frac{1/3}{x-1}dx\]

Answer 25

We might might to factor out the 1/3 :P might make it easier to read.\[\Large \frac{1}{3}\int\limits \frac{1}{x-1}-\frac{1}{x+2}\;dx\]

Answer 26

Good!

Answer 27

lol i said might like 3 times in that sentence XD silly typo...

Answer 28

Hahahah! Lol

Answer 29

Do you understand how to integrate it from there?
What types of functions should we end up with from those two bad boys?

Answer 30

So, the final answer is \[\frac{ 1 }{ 3 }\ln |x-1| - \ln |x+2|\]

Answer 31

But, we can simplify that!

Answer 32

Yes, you might want to put brackets to show the 1/3 is multiplying both terms though.
Yes, simplify! :)

Answer 33

Ugh, OpenStudy is tweaking out again -_-

Answer 34

Can you please show the simplification for this answer? Is it using the Log Rule?

Answer 35

Yes, log rule.\[\Large \color{royalblue}{\log(a)-\log(b) \quad=\quad \log\left(\frac{a}{b}\right)}\]

Answer 36

\[\Large \frac{1}{3}\left(\ln |x-1| - \ln |x+2|\right) \quad=\quad \frac{1}{3}\ln\left|\frac{x-1}{x+2}\right|\]
Sorry the site is being fussy again >:O grrr
You get something like this, yes? :O

Answer 37

It's fine! Don't worry! That is exactly what I got!

Answer 38

You could use another rule of logs to bring the 1/3 in as an exponent.
But I don't think you want to do that.
It looks better like this.

Answer 39

Yes, it looks fine like that!

Answer 40

I have to leave for a lil bit D: soz

Answer 41

It's fine! But, you will be back, right?

Answer 42

ya in like 45 mins prolly

Answer 43

Okay, perfect! I'll be here by that time! Thank you! Numbers 2 an 3 are really complicated!

Answer 44

Yessssssssssss! I got my new keyboard in the mail today!! :) So excited!!

Answer 45

http://www.duckychannel.com.tw/en/Shine_3_DK9008.html

So beautiful! :OOO

Answer 46

AWESOME!!!

Answer 47

That is so cool! :-)

Answer 48

Was super expensive :O But I've been dying to get a mechanical keyboard for a while now :3

Answer 49

ok ok ok back to the math XD

Answer 50

Hahahah! Okay, let's go back to math!

Answer 51

\[\Large \int\limits\limits \frac{ x^2+x-16 }{ (x+1)(x-3)^2 }\;dx\]Hmm so what's going on with the numerator there? :O does that factor?

Answer 52

Hmm doesn't appear to factor, right?
So we'll just leave it alone I guess.

Answer 53

This one is a tad trickier. We have a `repeated` linear factor in the denominator.
(x+1)(x-3)(x-3).
See the repeat? :O
So we have to do something a little sneakier with our setup.\[\Large \frac{x^2+x-16}{(x+1)(x-3)^2} \quad=\quad \frac{A}{x+1}+\frac{B}{x-3}+\frac{C}{(x-3)^2}\]

Answer 54

So far, so good! :-)

Answer 55

I'm not exactly sure how to explain the setup... I never really got a good explanation as to WHY we do it like that.
But you have to account for each copy of the linear factor.

For example if it had been (x-3)^3, we would need ANOTHER fraction, +D/(x-3)^3.

Answer 56

Yes! So far, so good! :-)

Answer 57

Ok so when we multiply through by our denominator we get:\[\Large x^2+x-16 \quad=\quad A(x-3)^2+B(x+1)(x-3)+C(x+1)\]

Answer 58

Okay, goo!

Answer 59

good!

Answer 60

gooo! there's goo everywhere! +_+

Answer 61

Remember earlier how we had gone over `2 methods` we could use from this point?
This problem is a really good example of one in which we REALLY want to use that second method (plugging in x values).

The other method:
Expanding everything and then getting a system of 3 equations, would be a ton of work.

Answer 62

So let's start by plugging in x=3.
Looks like that should cancel out 2 of the variables for us, yes?

Answer 63

Yes!

Answer 64

good good good :O so what you get for C?

Answer 65

1 sec

Answer 66

C = -1, right?

Answer 67

ya sounds good.

Answer 68

Hmm looks like we can get the A value by plugging in x=-1

Answer 69

A = -1/4, right?

Answer 70

Sorry was fiddling with keyboard +_+

Hmm I think A=1

Answer 71

On the left, the 1 and -1 cancel out.\[\Large -16 \quad=\quad A(-1-3)^2\]

Answer 72

Oh, yeah! I got it! That's right!

Answer 73

So we don't have a nice value we can plug in to solve for B.
So let's just plug in something simple like x=0.
We'll plug in our A and C values at this point also.

Answer 74

\[\Large 0^2+0-16 \quad=\quad 1(0-3)^2+B(0+1)(0-3)-1(0+1)\]

Answer 75

Got it!

Answer 76

B = 8, right?

Answer 77

mmm yah that sounds good.

Answer 78

Got it!

Answer 79

Still there?

Answer 80

@zepdrix

Outstanding work and patience.

Answer 81

:o

Answer 82

So I guess now that we've found our A, B and C, we should plug them in and the integrate, yah? :D\[\large \int\limits \frac{x^2+x-16}{(x+1)(x-3)^2}dx \quad=\quad \int\limits \frac{A}{x+1}+\frac{B}{x-3}+\frac{C}{(x-3)^2}dx\]

Answer 83

I got this random answer form the integration!

Answer 84

I got \[-\ln |x+1|+8\ln |x-3|+\frac{ 1 }{ x-3 } + C\]

Answer 85

Mmm I think you have your A and C plugged in backwards.

Answer 86

Oh, so it should be \[\ln |x+1|...\]

Answer 87

Wait, both A and C = -1!

Answer 88

Bahhh A=-1, crap my bad.
Was trying to do that step in my head :P That's what I get.

Answer 89

What do you mean you got a `random thing` from integrating XD lol
Looks like you got the correct answer :D

Answer 90

So, the first answer I wrote is right?

Answer 91

mmmmmmm ya i think so

Answer 92

So, is it that fine, or can we simplify it more?

Answer 93

Mmmm you can combine the 2 logs if you want I suppose.

Applying a rule of logs gives us:\[\Large -\ln |x+1|+8\ln |x-3| \quad=\quad \ln\frac{1}{(x+1)}+\ln(x-3)^8\]

Answer 94

\[\Large= \ln\frac{(x-3)^8}{|x+1|}\]

Answer 95

But I dunno, that doesn't really seem necessary :\

Answer 96

Ok too much math +_+ I need a break for a bit lol

Answer 97

Okay, perfect! Don't worry! Take your time!

Answer 98

You are back!