Original subtitles

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You're going to be talking about how you use partial fractions to evaluate an indefinite integral.

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And in this particular problem we've been given the integral of DX divided by the quantity x squared

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plus X and we've been asked to use partial fractions to evaluate this integral.

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Now as you get more and more practice with evaluating integrals it'll become more and more clear to

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you which integrals are good for partial fractions which are good for integration by parts and which

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are good for you substitution or maybe even a different method if you need any method other than just

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evaluating the integral directly one give away for partial fractions is you're looking for a rational

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function.

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Remember that by a rational function we mean a fraction where both the numerator and denominator are

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polynomials.

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And in this case we could take the DX out of our Fraction Right.

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DX is just essentially multiply by 1 over x squared plus x.

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So we have D X here and now it's really clear that we have a rational function because one is this constant

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value and x squared plus x.

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Both of these functions here are polynomials and we have one polynomial in the numerator one in the

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denominator.

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That's a rational function.

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So you might go right to partial fractions as a way to evaluate this integral.

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When you first look at it remember that partial fractions is just another way to simplify an integral

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that you can't evaluate directly in the same way that you can use integration by parts and you substitution

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to simplify an integral you can't evaluate directly the way that partial fractions works.

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To simplify these integrals we're going to basically decompose this fraction into smaller more manageable

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fractions.

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That's what it allows us to do.

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And it makes sense with the name partial fractions.

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Basically we're going to split this fraction into different partial fractions that together would make

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up this singular fraction but which separately are easier for us to evaluate So we're just trying to

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figure out how we can pull it apart.

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The person we want to do is make sure that we have a proper function.

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You always need to do that before you start your partial fractions decomposition.

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And what we mean by a proper function is the degree of the denominator must be greater than the degree

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of the numerator.

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Remember that the degree of a function like this.

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We talk about as the largest exponent on whatever you know x variable we have here.

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So in our denominator year we have x squared and essentially X to the first power here the largest exponent

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there is this too.

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So the degree of our denominator is to it's a second degree polynomial here in the denominator in the

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numerator.

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We don't even have an x variable so it's essentially X to the zero power or a 0 degree polynomial in

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the numerator.

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So in this case the degree of the denominator is larger than the degree of the numerator and therefore

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it.

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This is a proper rational function and we can proceed with our partial fractions see composition.

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If you don't have that situation you may need to do one of two things.

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Either make some kind of rationalizing substitution or perform a polynomial long division and I have

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videos about both of those methods so if you have that situation where your function is not proper then

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check out one of those videos to learn more about how to make it proper.

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But once we have it proper what we want to do is factor the denominator as completely as possible.

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We're going to take this function outside of the integral for now and we're going to factor the denominator

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as much as we can.

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So the denominator here we have x squared plus X we can factor out an X from that and what we'll get

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is X times x plus 1.

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And you want to factor as completely as you can.

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The more complicated here your polynomial may be the more factors you can have.

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But you want to get to the point where your factors are irreducible you can't factor them any further

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than you already have and that's what we've got here now at this point you want to look at the factors

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that you have and your denominator and there's a couple situations that can arise here.

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You can have either linear and or quadratic factors a linear factor is a factor whose degree is one.

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Right here we have two factors.

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We have X and we have X plus 1 in our factor here X we have X to the first power and an R factor X plus

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1.

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We have X to the first plus 1.

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These are both first degree factors and therefore their linear factors.

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If we had X squared for this first factor here that would be a quadratic factor anything x squared x

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cubed X to the fourth anything greater than 1 is a quadratic factor.

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And I have videos about how to do partial fractions decomposition with quadratic factors so you can

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check that out in a different video.

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Here we're only going to talk about linear factors.

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You can also have one of two situations with either of those linear or quadratic factors.

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You can have distinct factors and or repeated fact what we mean by that is if we were to have for example

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if our denominator was X times x plus 1 squared then this X plus 1 factor here would be a repeated factor

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because essentially our denominator would be x times the quantity x plus 1 times the quantity x plus

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1 because we had quantity x plus 1 squared.

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And you can see X plus 1 here is repeated.

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So we would call that a repeated factor whereas this x factor here.

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This factor only occurs one time.

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So we call that a distinct factor.

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I have other videos about how to deal with repeated factors so if you ever needed factors go check out

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one of those.

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But essentially what you can see here is that we're dealing with distinct linear factors.

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You can have distinct linear factors distinct quadratic factors repeated linear factors and repeated

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quadratic factors so all of those situations are going to be slightly different.

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Here we're dealing with distinct linear factors.

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So in this case when we have distinct linear factors we're going to set this equal to our partial fractions

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the composition on the right hand side.

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And when you have a linear factor you just want to put a single variable A as the numerator of that

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fraction.

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So we're going to have a over our first linear factor X and for our second factor we're going to use

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the next variable which would be B.

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So we're going to have B over our next linear factor X plus 1.

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Now a and b are going to change if you have quadratic factors if you have repeated factors you're going

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to need to change this a little bit.

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But when it's just linear factors and they're all distinct from one another they're not repeat in you

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just take each factor you put it as the denominator of its own fraction and you put a variable in the

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numerator of each fraction and you never repeat these variables so if we had a third factor we'd have

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another fraction you know plus C over X plus 2 or something like that.

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So this is distinct linear factors and once you have it set up like this here's what we're going to

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do and this is where things start to get a little interesting.

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We're going to multiply both sides of this function by the denominator from the left hand side x times

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the quantity x plus 1 x times the quantity x plus 1.

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When we do that we're going to multiply each term on both the left and right hand side by this value

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that's going to get on the left hand side.

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This denominator to cancel completely and all we're left with is 1 in the numerator over here on the

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left hand side on the right hand side when we multiply X times x plus 1 by this over X. We're going

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to get this x denominator here to cancel with this x and we're just left with a times x plus 1 for here

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b over X plus 1 the X plus 1 is going to cancel with this X plus 1 term.

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And we're just left with B times x.

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Once we've done this we want to go ahead and multiply out everything on the right hand side completely.

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So we'll have a x plus a plus b x.

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When we distribute that a across the X plus 1 then we want to group terms that have the same degree

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together.

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So we have here X to the first power and x to the first power which means those are going to go together.

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All of our constants if we have just a lone A or a loan B are constants are going to go together.

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If we had X squared values over here on the right they would all go together so we're going to say one

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is equal to.

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And we'll put them in parentheses so x plus be X like this and then plus a is our only constant so it

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goes over here by itself.

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Once you've done that you want to factor the X out from each of these values here so we'll say one is

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equal to.

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We're going to factor the X out of this and we'll get a plus b times X. We just pulled the X out and

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then we have plus a.

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Now the reason that we're doing this is because we're going to end up comparing coefficients on the

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left and right hand sides to create simultaneous equations that we can use to solve for our constants

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A and B.

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So what we have to do now is equate coefficient on the left hand side to those on the right hand side.

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And we have to think of our left hand side over here as 0 x plus 1.

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Right we haven't changed anything because 0 times X is 0.

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So we have to add anything actually to the left hand side.

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What we've done though is allowed ourselves the ability to identify that the coefficient of x on the

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left hand side is zero.

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The coefficient of x on the right hand side is a plus b right here.

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And then we equate our constants the constant on the left hand side is this positive one constant on

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the right hand side is this positive a right here.

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So what we do now is we equate those so we say zero from the left hand side is equal to a plus b and

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we say that one from the left hand side is equal to a.

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Now we have two equations and we can use the two of these to solve for our constants A and B Well conveniently

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We already have a value for a we know that a is equal to one from the second equation here.

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Now we can go ahead and plug in one for a to this other equation so we'll get zero is equal to a plug

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in 1 for a.

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And then we'll get plus B when we subtract 1 from both sides.

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We can see that we get B is equal to negative 1.

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And now we have a value for both a and b we've solved for our constants at this point we plug them back

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in to our partial fractions decomposition.

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This decomposition rate here that we had from before and we're going to be taking the integral of this

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decomposition instead of our original integral that we started with.

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So we're going to get the integral of a which we know is positive 1 divided by X from our decomposition

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before then we have plus B but we know that B is negative ones who are actually going to get minus 1

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divided by this original denominator here X plus 1 from our decomposition.

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And this is our new integral and as you can see these two fractions individually are much simpler to

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integrate than our original.

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Fraction one over x squared plus X and that's y partial fractions is so useful.

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So taking the integral of both of these terms individually you know that the integral of one over X

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is natural log of the absolute value of x or any natural log the absolute value of x.

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Then here for our second fraction we'll get minus because we have this negative sign here minus natural

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log of the absolute value of our denominator X plus 1.

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But then we need to apply chain rule and divide by the derivative of this X plus one value here.

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Well the derivative of X plus 1 is just one.

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So we divide by 1 it doesn't really change anything.

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Then we just need to remember to add c to account for the constant of integration.

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At this point we have our final answer except that we want to simplify as much as we can remember that

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we have a property of natural locks that tells us that natural log of a minus natural log of B is equal

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to natural log of a divided by B and that's the situation we have here.

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So instead of leaving the separate we can combine them and say that the simplification of our final

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answer is the natural log of X divided by X plus 1 and we keep our absolute value bars because we have

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those on both natural logs plus seeing is the integral of our original function.

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So that's it.

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That's our final answer.

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That's how you use partial fractions to evaluate an indefinite integral.