Partial Fractions

Start   sorry i wasnt here friday so uhh i guess last time we did some partial fractions but maybe not enough is that right?
alright so let me just do a couple of examples.
so the situation is where we have something we have something like
0:35something like this oh no this ones easy isnt it?
oh well say we have something like this anyway umm yeah that was a stupid one oh well.
if we have something like this then what do we do?
1:16you get something like the integral of..
1:50okay so lets do that.
oh no lets add a c.
2:10most of the people liked this option.
something like 75%.
and then this is number 2 not too many people liked this one.
in fact all of them work.
its just that if you do the substitution
2:31so lets do the substitution first so here if i make the substitution obviously i want to let u=x^2-1. i dont know if thats obvious to you but thats completely obvious to me.
and then du is 2x dx and i almost have an x dx sitting around there. i have that 3
3:02but so what?
so if i do that then i have x-3/x^2-1 dx then this will be i put this part and i have the 3 left over.
so actually let me leave that in.
3:30so i have that.
and now this i can do easily with this substitution so this is 1/2 u cause an x dx is a 1/2 du.
so i have a 1/2 du on the top and a u on the bottom and then here umm.
4:04well i guess this does matter right?
because now here i have to do do this 3/x^2-1 this is not 1.
i mean now i can make a trig substitution i suppose.
but i dont want to do this.
to do this 1/x^2-1 i have to make this substitution of sec^2
4:30i dont know, this is ugly.
is this ugly?
right? this i would have to do by a trig substitution.
so i dont really wanna go there.
is this ugly or am i just asleep?
its ugly right?
so i make a secant and i get a sec^2
5:01integral of well thats cos^2 then i have to do 1/2cos yeah?
so if i do 3/u then i have left over i need an x.
so i need some x here.
to change du/dx this would be 3/....du/2x.
5:39thats not so good.
so i cant just plug in the u.
because i have to deal with the dx.
so this i would just leave and now im kind of stuck i mean here if you have this you can do this
6:00so i can use a trig substitution thats sec^2 uhh x and now i get the integral of the cos^2 and then i turn that integral into 1/cos^2 with a half angle formula it works but actually maybe i shouldnt be sad this is sweat.
6:33soo anyway you can do it and itll eventually get there but its not the most efficient math.
soo formally the correct answer is either A or B and C together and depending on how much of a masochist you are it either doesnt matter or it does.
it matters in the sense that youre going to kill yourself
7:00or do something thats relatively easy.
so lets do it the easy way now so the way that most of you want to do is we just use partial fractions. so to remind you i just need to solve a little algebra problem A/x+1 + B /x+1 equals... x-3/(x+1)(x-1).
7:32so i want to solve this little algebra problem and figure out what A and B are and i can either do this by cross multiplying and getting 2 equations yeah let me do that. so this is the same thing as if A(x+1) + B(x-1)=x-3.
8:02right? cause if i find a common denominator and blah blah blah i get that.
and now here i can either say alright looking at this A+B has to equal 1 because the coefficients match. so im going to do this 2 ways.
so im going to do it both ways, you choose which way you like.
theyre the same to me.
8:30so either i could say alright if i clean this out i have Ax + A but im going to not worry about that im going to leave the x off of that. A + B has to equal 1 because the powers-the coefficients of x A+ B have to match.
and then the coefficients of the constants A-B have to = -3.
9:02so this is from the x and now i solve..thats a 3?
funny looking 3.
and now i solve those simultaneously.
so if i add those things together i get that 2A = -2.
and so A
9:32= -1.
and since A+B is 1 that means B is 2.
so thats 1 way i can do it.
the equivalent way i can do it is i can look at this equation and i can say okay this equation has to be true for all x's.
so in particular this equation has to be true
10:00this equation has to be true when x is 1.
if x is 1 then this is 0 and its gone and its out of my face.
so that means that A(1+1) so thats 2A = 1-3.
and so that means
10:30again A is 1.
cause 2A has to = -2.
and if x=-1 which kills this piece then i have here -2B = -1-3 -4 so that means B is 2.
11:00these are the same just done in a slightly different way.
i mean whats making them work is exactly the same facts which is that this equation has to hold for all values of x.
so either you can equate coefficients or you can pick suitable values of x and whichever one makes you happy is fine.
theyre really the same.
alright so in both cases
11:33in both cases we wind up with something fairly easy.
so that tells me that A is -1 and B is 2.
that means that the integral x-3 dx over (x+1)(x-1) is the same thing as the integral of
12:01-dx/x-1 + the integral of 2dx over x+1 and this is really easy.
right? so this is look at it if i make the substitution u=x-1 then i get 1/u this is -ln
12:31of x-1 and here i make the substitution u=x+1 so i get 1/u so this is +2 ln x+1.
so these are really easy once you get the hang of them.
the hard part so the hard part with partial fractions actually as in most calculus is not the calculus but the algebra.
13:02but this is pretty straightforward.
any questions on this at all?
then why did you make me do it?
okay soo then lets just go on.
umm let me do another one. so unfortunately i think this is what you get at he end of the class what she told me confused you all umm maybe all the factors are not linear.
13:31maybe i have a power on the top or the bottom. or something like that.
and i cant factor it into linear factors cause i cant always factor everything into linear factors.
so the problem here is well let me just do an example still.
so say i have the integral of 3x^2-x...gotta work these out in advance
14:04+5 over (x-2)^2 +1 and theres a dx here somewhere.
and again it should be pretty obvious that what i really want to do here is partial fractions because umm
14:36so now i want to be able to split this up.
the problem is so let me actually move over to the next board.
so i have this problem 3x^2=x+5 (x-2)^2(x+1)
15:00and i want this to be equal to something over (x-2)^2 + im going to skip B over x+1 but this wont actually work.
because im going to be missing an x.
this only gives me a linear term in x and it doesnt give equal choice on the x term.
an equal choice on the constant term. the constant term and the x term are related by that.
15:34so i need a slightly more complicated numerator here.
so that i have the full freedom to vary the coefficient of x.
well this is really the coefficient of x^2 and the coefficient of x.
because when i cross multiply this one gets a quadratic term this one gets a linear term and loses out.
on the freedom. i need 3
16:00things to vary.
because i have a quadratic polynomial or a cubic one so i need 3 degrees of freedom.
the general rule is or atleast some people say it slightly differently when you take your factors you always put the numerator so the numerator is always a polynomial
16:341 degree lower than the denominator and then maybe i split this up.
those of you that may have seen this before or the way they describe it in the book they say okay take A/(x-2)^2 + B/x-2
17:00+ c/x+1 those are the same just slightly different choices of A B and C.
the advantage of doing this is this unifies the 2 second cases.
any questions on that at all?
student: does it matter which letter is assigned the x value?
no. would you like to make this be a B and this be a theta?
it doesnt matter. now you cant read it so thats probably a bad idea.
17:34the letter we can put smiley face here and a star here.
it doesnt matter because its just a place holder.
so use whatever letter you want.
but then once you start be consistent.
you understand? so what matters is that the thing on top of the quadratic term
18:00is a linear polynomial. and the thing on top of the linear term is a constant.
should i leave the smiley face and star? yeah.
okay if you like smiley face and star we'll keep it alright so now we can use the same trick you just cross multiply and come up with something.
so when we cross multiply we're going to get smiley face x + star
18:32times x+1 +c times (x-2)^2 and that has to equal that 3 x^2-x+5.
and so now we need to figure out what smiley face star and c are.
and again we can use either method.
19:01lets just equate coefficients this time. so if i equate co-well... yeah lets equate coefficients.
cause otherwise i'll have to choose 2 values for x.
so here if i multiply this junk out i get smiley face x^2 plus uhh smiley face plus star.
+ star plus so im just distributing this.
is it confusing cause im not using actual letters?
you could no ahhh ughhhhh!
ok sorry.
umm and then here i have to square this out.
so this squares out to be x^2-4x+4.
20:01and when i multiply that by c get that. so this is cx^2-4cx+4c and that has to equal 3x^-x+5.
and so from there
20:31now we can match things. so the x^2's give me a 3 so i have smiley +c =3 and the x's tell me that smiley + star -4c =-1.
21:07and then from the units terms i get a star a +4c and spot.
and so now i have to solve these things all together.
21:37so if i add these together so if i add these together i get.. so adding this one and this one will give me smiley face + 2 stars =4.
22:02and.. uhh did i make a mistake?
somebodys telling me i made a mistake.
yeah you had a question.
okay so lets go through this process again.
i look at this equation here i cross multiply everything i get some long mess.
22:30now what has to be true this is whats in front of an x^2?
in front of an x^2 i have 1 smiley face i got a c and i have a 3.
and this is whats in front of an x.
in front of an x i have a smiley added to a star i have a -4c i have a -1.
23:02and this is whats in front of a 1.
in front of the 1 here 1 has an x blah blah blah in front of the 1 here i have a star.
and i have a 4c.
and on this side i have a 5.
so that gives me my 3 equations.
and now i just want to say if i have 3 numbers that when i add up in this way i have 3 numbers and they have to satisfy these 3 equations.
23:32now i want to solve these 3 equations.
maybe i should do it the other way.
if i do it the other way it'll help me see right away.
so lets look at the other way so you could solve these 3 equations you'll probably get the same answer i will the other way.
the other thing i could do is i could say this has to be true for all x's.
if x is 1
24:00this mess here is 0.
x is -1 then i have spot smiley plus star times 0 plus c times (-1-2)^2 = uhh some number 3 +1 +5.
24:31so this is gone and so this is 9 so i get 9c = 9 well thats good.
so that tells me right away that c is 1.
now that i know what c is then i can go back and figure out what smiley and star have to be.
if i take x to be 2 well that gives me a relationship between smiley and star so actually im going to mix these 2 methods.
25:04so i know that c is 1 which makes my life a whole lot easier cause i know that smiley smiley + 1 is 3 from this line here.
well since smiley + 1 is 3 that means smiley is 2.
25:31and since smiley is 2 and here i know that c is 1 so star + 4c is 5 star + 4c is 5 so that means that star is 1.
so now mixing the 2 methods actually turned out to be the most efficient way.
you can of course solve these by adding subtracting and mushing them around it works
26:00and you can of course use this method exclusively as well.
so let me finish this and then ill address that.
so are we okay on this?
26:33so use whatever method works as long as its legal.
thats fine. im mixing the 2 methods theyre the same method so it doesnt matter.
this all comes down to the fact that if 2 polynomials are equal they have to be equal at every x value and their coefficients have to be equal.
if 2 things are equal they have to be equal so that means they have to be the same.
so i can use their various aspects to identify them.
if youre trying to identify your mother
27:02you can listen to her voice or look at her face theyre the same.
theyre both ways of identifying who your mother is.
okay soo anyway so we have this and so now this problem becomes easy somewhere.
i guess it'll be easy over here.
so that means that the integral of 3x^2
27:30-x+5 dx over..whatever all that junk is (x-2)^2(x+1) is the same thing as the integral smiley is 2 2x+1 over (x-2)^2 dx
28:00and c is 1 dx over x+1 and then this guy you can do by substitution and this guy is the log so we go and we stop there.
so let me just not do that.
now let me come back to somebody over here.
28:32pink and the blue right?
so her comment about why dont i use d instead.
because its the same.
uhh let me use d on that board.
so if you like if it makes you happy
29:03i could write this a different way i could say instead that 3x^2-x+5 (x-2)^2(x+1) is A/(x-2)^2
29:31B/x-2 C/x+1. i could do that too.
is that what you meant?
forget it?
so i can do this as well, its the same.
because if i put these things together then my B will just be some combination of smileys and stars.
right? because
30:08so A is the same as smiley and B is a smiley and a star which is B+A...B-2A is a star.
oh yeah yeah.
30:34but its still the same thing morally. so the A is the star and i mean B is the smiley and A-2B is star.
so that doesnt matter.
and if you want to put more stuff up here if you want to put D's and E's and Q's times x's and x^3's its fine but theyll all be 0's.
31:00so why bother?
so that-yeah?
what else would it be?
okay so if i tried to do x-2 and x+2 then i wouldnt have a way of you mean you would like this one to be like that?
31:32is that what youre asking?
so this is a good question and its something that confuses people all the time.
imagine that i did that so imagine that i wrote a well okay so lets do a simpler example.
suppose i did that.
suppose i wrote A/x-2 + B/x-2
32:00how can i get an x^2 out of that?
you cant.
but that has to equal something with an x^2.
right? cause i have an x^2 on the top.
i cant get an x^2 out of that because when i cross multiply im going to have A(x-2) + B(x-2).
theyre not going to get squared.
32:30so its not going to happen.
so i need to have something squared.
so i could square this and then when i cross multiply the B will get the x^2 term.
somehow im going to have to pick up the x^2 so the general situation is that whenver i have something
33:00over factors whenver i break this down the thing on top is always the polynomial of just 1 degree lower of the thing on the bottom.
now sometimes you might have to like in this example actually you might have to split it up.
because when i make my substitition its not going to quite work but i can split that up.
33:37so if i have the integral of (x-2)^3 times x over i dont know, an x.
so i want to split this up.
and im going to write this as A/(x-2) cubed i can either write it as 3 things where i do a 2 a 3 a 1
34:02or i can just write it as 1 thing like that.
and since i put an integral there ive got one right here. yeah?
so okay i am solving by elimination yes
34:31so i can solve this by elimination but i have extra information that i might as well use.
i dont like solving 3 equations and 3 unknowns because i combine this and i combine that and now i get 2 equations and 2 unknowns then i solve the 2 equations and 2 unknowns and then i go back and find the others its easy for me to figure out what c is not using this.
so by looking at the equation here
35:01that if x is -1 this term is gone.
when the x is -1 this thing is 0.
and then i have 3c -9c =9.
so by the judicious choice of x i can make the equation simpler.
35:39im saying sometimes 1 method is faster than another.
so what i dont like to do is say always use this method because its always best.
its not true.
sometimes one methods better sometimes another methods better sometimes a mix of the 2 methods are better.
there are 2 methods that are the same really
36:01they seem different and they have a different feel and you use whichever works best at the time.
you can- i can do this i can forget about the other method entirely.
and just solve this by itself.
so if youre confused by the other method forget about it.
just solve the 3 equations and the 3 unknowns youll get the same answer i guarantee it.
but its slow, it sucks.
36:30but youre welcome to do it.
this other method isnt going to work so well here because i can kill the x^2-2 by substituting x=2 but then im left with something involving both smiley and star so i need to make another choice that will not kill all 3 terms and put that together.
so instead of doing that i mix these 2 methods.
theyre all the same thing so you can do it.
37:04umm okay other questions about this?
we will-- I mean it may seem like-- I mean this notion of partial fractions comes back outside of doing integrals; it still comes up in other areas.
so let me stop with this guy.
37:30let me point out one other thing that will be more algebra ---so really today is an algebra class--- suppose that i have something like let me just do some integral The next one is x^3-4^2x+1 over x^2-1
38:03so let me forget about the integration part so suppose i want to do this what should i do?
should i write this as ....
38:36so A yes B no i seem to have gotten a lot of people so i mean ill stop there unless you need more wait, okay.
so by overwhelming majority you think this wont work.
39:00and youre right why wont this work?
i cant get an x^3 so what do i do?
i do what?
i divide it.
does come from the denominator i mean i cant start with partial fractions id do something first.
39:33so the problem is that the power here is 3 and the power here is 2 right if you prefer i could write this as x^2-1 The power here is 3 and the power here is 2 partial fractions is not going to work i have to clean it up a little first.
so this is the same thing
40:06and so what i have to do is reduce the power of the numerator.
so i want to get that down to being an x^2.
or yeah soo i mean actually so i need to do division i need to say if i multiply the bottom by x ill get x^3 that i can subtract off
40:30so i have to do long division here.
i dont have to do it exactly in this form but okay.
guess ill leave room for x.
and probably a lot of you forgot how to do long division with actual numbers. how many people remember how to do that?
if i want to do like 15 and 20961 people remember how to do this from like 3rd grade?
41:03it works exactly the same.
so when im dividing 15 into 2096 i say okay well 1 goes into 2..2 times no thats too big so i use it 1 time so i put a 15 here i put 1 here i subtract and i say how many times does 15 go into 59 uhh 5 times no uhh 4 times no almost. 3 times and then i get 45
41:303 i subtract and i get a 14 6 and i say 15 into here maybe 9 no and so on and i just do this 9?
9..45..youre right.
okay so 9 works and when i do 9 into here i get uhh 90 +45 is 135
42:02im already tired but now its so close i should finish right?
so if i go 15 into here it goes 4 times..5 times.. 7?
7 is too big. 6.
so if i do 6 then i get 60+30 is 90 7.
105 and i have a remainder of 6.
42:31im actually going to write it as a remainder of 6 rather than writing 6/15.
but if you prefer you can write it as 6/15 okay here the trick is exactly the same.
i look at the x^2 i say how many times does it go into x^3?
it goes in exactly x times and so if i whoops what am i doing?
so it goes in x^2 goes into x^3 exactly x times
43:03and then x times x^2-1 is x^3 -x.
and i subtract which is exactly what i did here 15 goes into 20 how many times? 1 time 1 times 15 is 15 then i subtract this part then i bring down whats left over.
so here i subtract the x^3 is gone now
43:31when i subtract here a 4..-4x^2 -x +1.
plus i subtracted the minus thank you.
and i can do this one more time i can say how many times will this x^2-1 go into -4x^2 blah blah blah well x^2 goes into 4x^2 -4 times so then i multiply back just like i did over there.
44:01-4 times x^2 is -4x^2 if this term doesnt cancel you screwed up.
and then -4 times -1 is +4 4 and then again i subtract those guys cancel x-0 is x and this guy i mean 1-4 is -3.
44:30so then then this problem becomes so that tells me oh i guess i need the remainder. this is my remainder.
if i try to divide x^2+1 into this i will need to divide by x rather than multiply by x so that means that this process has showed me that
45:01that x^3-4x squared + 1 over (x+1)(x-1) is the same thing as x-4 + x-3 over
45:30(x+1)(x-1) and so if im integrating that so if i want to integrate this then i integrate this and i integrate this this one i did already at the start of the class.
so i choose this on purpose to make sure it would come out so now i can do this by partial fractions.
46:06so i will not do this by partial fractions because i already did it.
right there.
46:32so if for partial fractions to work you need the degree of the numerator to be 1 less than the degree of the denominator.
because otherwise you cant pick up the extra term.
so sometimes you have to rearrange a little.
okay so umm let me finish this with uhh thats pretty much it for
47:00techniques of doing integration so you now know as much as i do about how to integrate but now theres some other variations on them which we'll focus on in the next few classes.