WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to be talking about a technique called integration by partial fractions.
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Let us jump right on in.
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I think I will stick with black, for the time being.
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Recall the concept of finding a common denominator.
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Recall the idea of finding a common denominator.
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Something like, if we had 2/ x + 2 - 3/ x – 3.
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We want to find the common denominator.
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The common denominator is going to be x + 2 × x – 3.
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What we end up doing, we multiply this, so we get 2 × x – 3.
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Nice, basic mathematics from long ago.
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3 × x + 2/ x + 2 × x – 3.
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Of course, we multiply everything out.
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We end up with 2x - 6 - 3x - 6/ x² 2x - x – 6.
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2x, we get - x - 12/ x² – x – 6.
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This is the same as that, except under a common denominator.
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Now what if we have to evaluate?
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What if we were asked to evaluate the following integral.
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The integral of – x - 12/ x² – x - 6 dx.
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If we had a procedure from going backward, from this to this, we can write this integral this way,
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and then just integrate both of these which we know how to do.
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What if we had a procedure from going backward from - x - 12/ x² - x - 6 to 2/ x + 2 - 3/ x – 3.
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Then, we would do the integral of 2/ x + 2 dx - the integral of 3/ x - 3 dx.
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We know what this equal already.
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We already know that this = 3 × natlog of x + 2 - 3 × natlog of x - 3 + c.
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What we want to do is we want to find a way of taking a rational function,
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working backward, breaking it up into what we call partial fractions.
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We want to do a partial fraction decomposition and then,
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we want to integrate each one of those partial fractions based on techniques that we have already developed,
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either straight integration or any of the other techniques that we have developed.
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That is what we want.
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This procedure, if we are going backward, it is what we will develop in this lesson.
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This procedure, if we are going backward is what we develop here.
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Some of you may have seen it, some of you perhaps you have not seen it.
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Not a problem, it is what we develop here.
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We call it decomposing a rational function.
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Rational just means you have a numerator which is a polynomial, the denominator is a polynomial.
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Rational function into its partial fractions.
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That is why we call it the partial fraction decomposition.
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There are two types of rational functions.
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An example of one is, let us say x³ + 5/ x² + 9.
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Another would be x² + 9/ x³ + 5.
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Here, this is called improper.
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The reason it is improper is because the degree of the numerator.
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The 3 is bigger than the degree of the denominator.
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The degree of the numerator is bigger than the highest degree of the denominator, that is called improper.
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This one, where the degree of the numerator is less than the degree of the denominator, that is called proper.
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Proper because the degree of numerator is less than the degree of the denominator.
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Any proper rational function can be decomposed into its partial fractions, any one.
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If the rational function that we have to deal with is already proper, it is fine,
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we just subject it to this procedure that we are going to develop to decompose it to its partial fractions.
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How do we deal with it, if we have a rational function which is not proper?
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We do a long division and we turn it into something proper.
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The question is, but how do we handle an improper rational function?
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We perform a long division, a polynomial long division, if you remember from algebra.
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We rewrite this numerator which is a function of x/ denominator which is a function of x.
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Something that is improper, we do the long division, and we get a quotient
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which is a function of x that is the one on top + the remainder that we get,
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which is the one in the bottom/ the divisor which is the denominator.
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We rewrite it, we do a long division and express it this way.
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We take the integral of this.
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The integral of this, whatever we need to do here, this now is going to be proper.
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We can subject that to a partial fraction decomposition.
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Here, this is improper, this is just our quotient.
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This is now going to be proper which we can then decompose.
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Proper, this, we decompose.
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Let us do our example.
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Let us have a rational function, let us go back to black.
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I have got a rational function f(x) is equal to x³ + 5/ x² + 5.
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This is improper, let us do the long division.
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The long division is x² + 5 x³ + 0 x², I have to fill in every space, + 0x + 5.
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What times x² gives me x³?
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It is going to be x.
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x × x² is going to be x³.
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x5, I’m going to put the 5x here.
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I'm going to change the sign.
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This cancels with that, I'm left with -5x + 5.
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This is my quotient, this is my remainder.
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This is my divisor, my denominator.
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We have this thing, turns into is equal to x + -5x + 5/ x² + 5.
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There you go, this is proper and I could subject it to a partial fraction decomposition which is the next thing we are going to do.
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That is it, anytime you have an improper, do the long division.
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You are going to get a quotient + a proper fraction.
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Let us go ahead and talk about our different case now.
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Let us say every proper rational function can be decomposed into its partial fraction sum.
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In other words, it can be expressed as a sum of partial fractions.
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Every proper one can, every proper fractions, where the denominators of the partial fractions,
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of the individual partial fractions are linear factors and/or irreducible quadratic factors.
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Let me explain what this means.
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You actually know what this means already, you have seen it a thousand times.
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Anytime you factor a polynomial, you can always factor it and get a bunch of factors multiplied by each other.
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Those factors are either going to be linear, in other words,
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the exponent of the x is going to be 1 or they are going to be quadratic.
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The exponent is going to be 2.
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Irreducible means you hit a quadratic function that you cannot factor any further.
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It is always going to be that.
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It is always going to be 1 or 2.
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You can always break it down.
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You are never going to have something that is going to be cubic, or quartet, or quintet.
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It is always going to be a linear factor which is x¹ or a quadratic factor x².
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Here is what they look like.
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A linear factor looks like this, partial fractions of linear factors.
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It is a fraction so it is going to have some numerator.
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A partial fraction, let us read this again.
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Every rational function can be decomposed into partial fractions sum, that can be expressed as a sum of partial fractions,
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where the denominators of the partial fractions are linear factors or irreducible quadratics.
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The linear factor is going to be some a which is a constant/ ax + b, to some m power.
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Linear factor looks like this.
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Linear means the power of x is 1, x + 6, 2x – 2, 3x + 15, these are linear factors.
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Quadratic factor looks like this.
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The exponent on the x itself, not this one up here.
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These are multiplicities, that is however many times the root shows up, when you factor a polynomial.
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A linear is when the exponent here is 1.
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Quadratic is when the exponent on the x itself is 2, it is a quadratic factor.
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Notice, this is a power of 1, the number on top has to be a constant.
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It has to be 1° less.
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A quadratic partial fraction is quadratic in the denominator, the x, it is linear in the numerator.
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It is always going to be like that.
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In your partial fractions, these things, however many there are.
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1, 2, 3, 4, 5, the denominator is linear, the top is just a number.
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If the denominator is quadratic, the top is some linear function.
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It is going to be bx + c.
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It is going to be 1° less.
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The letters are just constants.
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Do not worry, everything will make a lot of sense in just a minute.
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It is actually very simple, it is quite algorithmic.
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There is no problem.
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We distinguish four different cases.
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That is fine, I will go ahead and do it this way.
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We distinguish four cases.
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In case 1, we have a rational function where, in other words f(x)/ g(x).
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Case 1 is when we factor the denominator which is what we are always going to do first, it takes a rational function.
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You are going to break down the denominator as much as possible.
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You are going to factor it out, as much as possible.
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G(x) is a product, if when you factor it, you find that it is a product of distinct linear factors,
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case 1, g(x) is a product of distinct linear factors.
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In other words, this means this, thru words.
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g(x), when it is factored is equal to some a1x + b1 × a2x + b2 × some a sub nx + b sub n.
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In other words, no factors are repeated.
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These a's and bs are different.
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a1 and b1, a2 and b2, an, bn, they are all different.
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In other words, it does not have a multiplicity of more than one, that is what this means.
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Distinct linear factors, it means a multiplicity that factor occurs only once.
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No factor is repeated.
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In this case, the partial fraction decomp looks like this.
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f(x)/d(x) is going to equal some a1/ a1x + b1 + a2/ a2x + b2 + so on and so forth, + a sub n/ a sub nx + b sub n.
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Our task, in other words, we are going to write it out like this.
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Our task is going to be find what a1 through an are.
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Our task is to find a1, a2, and so on, all the way to an.
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All we have done is factor the denominator into its factors.
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Written those factors, we know that it breaks up into some composition of partial fractions.
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We need to find what the numerators of those partial fractions are.
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An example will make this clear.
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Let us work in blue.
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Find the integral of x² + 3x - 10/ 3x³ + 8x² - 3x.
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The first thing we always do is factor the denominator, that is what we want.
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We want to break it up into its factors.
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They are going to be either linear or they are going to be quadratic.
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We factor as far as we can.
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We always start by factoring the denominator.
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If it is already factored, you are done, that part is taken care for you.
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All the factors will be written.
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Sometimes it is given that way.
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We always start by factoring the denominator, as far as possible.
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Again, the factoring will always be linear, quadratic.
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You might be only linear, you might be only quadratic.
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You might combination linear and quadratic, but it is always going to be linear or quadratic.
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This first case that we are dealing with is they are all linear and they are all distinct, multiplicity 1.
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Let us take our denominator which is 3x³.
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We have got 3x³ + 8x² - 3x.
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This is equal to x × 3x² + 8x – 3, that is equal to x × 3x - 1 × x + 3.
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There you go, that is our full factorization.
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Linear, linear, linear, 3 factors.
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Linear, exponent is 1, linear, exponent is 1.
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They are all distinct.
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This, this, and this are completely separate, they are completely distinct.
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We have three linear factors.
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Our partial fraction decomp looks like this.
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It looks like, we have x² + 3x - 10/ 3x³ + 8x² - 3x is equal to, it is equal to some a,
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some constant a/ the first factor x + some constant b/ the second factor 3x - 1 + some constant c/ the third factor x + 3.
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That is what we meant.
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This is a common denominator.
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This, this, this, this and this are actually the same.
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I know that there are three fractions, that when I combine them to get a common denominator, I'm going to get that.
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I'm working backward.
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My task is to find a, find b, find c.
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There are three distinct linear factors.
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I put each one of the denominator as a separate fraction, a, b, c.
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I need to find a, b, c.
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We want to find a, b, and c, that is what we want to do.
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Let us go ahead and do that.
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This is how we start, we write out the decomposition.
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What we are going to do, this is equal to that.
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I need to find this, this, this.
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What I'm going to do is I'm going to express the right side, in terms of the common denominator.
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The common denominator, I know what the common denominator is.
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The common denominator is this × this × this.
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That means I'm going to multiply a by this and this.
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I’m going to multiply b by this and this.
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I’m going to multiply c by this and this.
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When I do that, you will see what happens when I do that.
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Now we write the right side.
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We write now the right side, in terms of a common denominator.
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We have x² + 3x - 10/ 3x³ + 8x² - 3x is equal to a × 3x - 1 × x + 3
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+ b × x × x + 3 + c × x × 3x - 1/ x × 3x - 1 × x + 3.
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This is equal to that.
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Since this is equal to that, we can sort of ignore them, that means the numerator is equal to the numerator.
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Let us expand the numerator and see what we get.
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When I expand the numerator, this is going to be a ×,
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that is fine, I will go ahead and just work with that.
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Let us go ahead and expand that.
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We have a × this is going to be 3x² + 8x - 3 + bx² + 3bx + 3cx² – cx.
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It is going to be 3ax² + 8ax – 3a + bx² + 3bx + 3cx² – cx.
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I’m going to combine terms, combine common terms.
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In other words, anyone that has x² and x² in it.
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Let me do this one in blue.
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I lost my little color changing thing.
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We are stuck with red for the rest of the time, not a problem.
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Combine the common terms, 3ax² and x².
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I have got x² × 3a, I’m going to pullout the coefficients.
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3a takes care of that one and x² is + b.
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I’m just combining common terms, by combining the coefficients 3a and b.
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I’m just writing it with a coefficient on this side, instead of the other side.
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I hope that is not too much of a problem.
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I have another one for x².
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I have got 3c, sorry about that.
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ab + 3c +, I have x terms, x.
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I have got 8a, it takes care of that one.
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+ 3b – c, that takes care of the x term.
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I have +, the only term I am left with is this -3a.
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I have expanded the numerator, now I have this thing.
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We have this, we now have left side which is x² + 3x - 10/ 3x³ + 8x² - 3x,
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all of that is equal to x² × 3a + b + 3c + x × 8a
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+ 3b - c + -3a/ x × 3x - 1 × x + 3.
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We expressed it, multiplied it out, now I have this.
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I will go back to red.
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The denominators are now the same, except one is non factor form and one was factored form, but they are the same thing.
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Because the denominators are the same thing, I can ignore them.
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That means the numerators are the same thing.
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That means this numerator is equal to that numerator.
00:31:07.600 --> 00:31:13.500
Because they are equal, every term on the left has a corresponding term on the right.
00:31:13.500 --> 00:31:17.400
Here we have an x² term, the coefficient is 1.
00:31:17.400 --> 00:31:24.400
That means over here, there is an x² term, that means this is its coefficient, it is equal to 1.
00:31:24.400 --> 00:31:27.700
This x term, its coefficient is 3.
00:31:27.700 --> 00:31:32.500
This x term, that is this thing, this is equal to 3.
00:31:32.500 --> 00:31:38.100
-10 is the number, -3a, this is equal to -10.
00:31:38.100 --> 00:31:42.000
I set equal the coefficients of corresponding terms.
00:31:42.000 --> 00:31:44.400
I get a system of three equations and three unknowns.
00:31:44.400 --> 00:31:47.100
I’m going to solve for a, b, and c.
00:31:47.100 --> 00:31:50.400
That is how I do this.
00:31:50.400 --> 00:31:54.400
Because the numerators are equal, now what I have got is the following.
00:31:54.400 --> 00:32:20.600
x² + 3x - 10 is equal to x² × 3a + b + 3c + x × 8a + 3b - c + -3a.
00:32:20.600 --> 00:32:34.000
The only way that these two are equal, I know they are equal if the denominators are equal.
00:32:34.000 --> 00:32:56.300
The only way left and right side are equal is if corresponding coefficients are equal.
00:32:56.300 --> 00:33:09.400
In other words, 3a + b + 3c has to equal 1.
00:33:09.400 --> 00:33:29.600
8a + 3b - c is equal to 3 - 3a is equal to -10.
00:33:29.600 --> 00:33:36.400
-3a – 10, that implies that a = 10/3.
00:33:36.400 --> 00:33:40.400
I have already found my a.
00:33:40.400 --> 00:33:43.700
I’m going to put that a into this equation.
00:33:43.700 --> 00:33:46.200
I’m going to put this a into this equation.
00:33:46.200 --> 00:33:51.900
I'm going to get two equations and two unknowns.
00:33:51.900 --> 00:33:53.400
I wonder if I should go through process.
00:33:53.400 --> 00:33:55.200
That is fine, this is probably a good review.
00:33:55.200 --> 00:33:56.800
I’m going to put a in here.
00:33:56.800 --> 00:34:07.500
This is going to be 3 × 10/3 + b + 3c is equal to 1.
00:34:07.500 --> 00:34:19.400
And then, I have 8 × 10/3 + 3b - c is equal to 3.
00:34:19.400 --> 00:34:31.000
This implies, when I multiply, move things over, I'm going to end up with b + 3c is equal to -9.
00:34:31.000 --> 00:34:48.000
I’m going to end up with 3b - c is going to equal -71/3.
00:34:48.000 --> 00:35:08.600
I'm going to multiply the top by -3, that gives me -3b - 9c = 27.
00:35:08.600 --> 00:35:09.500
I will leave the other one alone.
00:35:09.500 --> 00:35:15.900
3b - c = -71/3.
00:35:15.900 --> 00:35:30.000
Add them straight, I end up with -10c is going to end up equaling 10/3, that means c is going to equal -1/3.
00:35:30.000 --> 00:35:36.300
I found my c, now I put that into one of these equations.
00:35:36.300 --> 00:35:48.800
I will put it into this one, I get 3b, - and -1/3 = -71/3.
00:35:48.800 --> 00:35:55.000
I get 3b + 1/3 = -71/3.
00:35:55.000 --> 00:36:05.900
I get b is equal to 8/9, when I solve.
00:36:05.900 --> 00:36:11.700
I found a, I found b, I found c.
00:36:11.700 --> 00:36:17.100
Now I put them back into my partial fraction decomposition that I actually wrote first.
00:36:17.100 --> 00:36:24.200
Let us go back.
00:36:24.200 --> 00:36:38.000
What happened here, I did something here.
00:36:38.000 --> 00:36:45.800
Now I have, remember the partial fraction composition that we,
00:36:45.800 --> 00:36:53.000
Something is wrong with my, let us try black.
00:36:53.000 --> 00:37:05.600
I have got x² + 3x - 10/, remember our original, 3x³ + 8x² - 3x.
00:37:05.600 --> 00:37:07.200
We wrote the partial fraction decomposition.
00:37:07.200 --> 00:37:20.000
We said that that is equal to a/x + b/ 3x - 1 + c/ x + 3, that was our partial fraction decomposition.
00:37:20.000 --> 00:37:22.400
We have manipulated this partial fraction decomposition.
00:37:22.400 --> 00:37:25.400
We set corresponding coefficients equal to each other.
00:37:25.400 --> 00:37:26.600
We solve the system of equations.
00:37:26.600 --> 00:37:28.700
We found a, b, and c.
00:37:28.700 --> 00:37:42.000
We found that it is equal to 10/3 / x + 8/9 / 3x - 1 + -1/3.
00:37:42.000 --> 00:37:43.200
We usually leave it like this.
00:37:43.200 --> 00:37:49.100
With + in between, we leave the - on top, / x + 3.
00:37:49.100 --> 00:37:53.600
This is our partial fraction decomposition of that.
00:37:53.600 --> 00:38:01.700
Our original was, what we started out doing, we wanted to find the integral of this.
00:38:01.700 --> 00:38:06.400
That is just the integral of this, we decomposed it.
00:38:06.400 --> 00:38:25.800
It is the integral of that, 10/3 / x + 8/9 / 3x - 1 + -1/3 / x + 3 dx.
00:38:25.800 --> 00:38:35.500
That is equal to 10/3 × the integral of 1/x dx + 8/9 × the integral of 1/ 3x - 1 dx -1/3
00:38:35.500 --> 00:38:54.900
× the integral of 1/ x + 3 dx this = 10/3 ×,
00:38:54.900 --> 00:38:59.400
we did the partial fraction decomposition because now these are all logarithms.
00:38:59.400 --> 00:39:00.600
That is why we did it.
00:39:00.600 --> 00:39:17.200
The natlog of the absolute value of x + 8/9 × 1/3 × natlog of 3x – 1.
00:39:17.200 --> 00:39:24.500
1/3 comes from the fact that this is 3x – 1, we use a u substitution real quickly.
00:39:24.500 --> 00:39:34.400
u = 3x – 1, du = 3dx, dx = du/3.
00:39:34.400 --> 00:39:38.800
Therefore, this integral is, the integral of 1/u du.
00:39:38.800 --> 00:39:42.700
du, the 1/3 comes out, that gets pulled out as a 1/3.
00:39:42.700 --> 00:39:52.600
This is -1/3 × the natlog of x + 3 + c.
00:39:52.600 --> 00:39:54.400
There is your partial fraction decomposition.
00:39:54.400 --> 00:40:03.300
This partial fraction decomposition are very long and they are very tedious.
00:40:03.300 --> 00:40:05.700
That is just the nature of the game.
00:40:05.700 --> 00:40:11.400
I would recommend using some online software, as far as solving the equations and unknowns
00:40:11.400 --> 00:40:14.400
because you might have three equations and three unknowns.
00:40:14.400 --> 00:40:17.100
You can work with it, sometimes, you may have 3, 4, 5.
00:40:17.100 --> 00:40:22.400
I would definitely just use some software, in general, at least to get through the problems.
00:40:22.400 --> 00:40:27.500
But you want to go through the partial fraction decomposition, to finding the common denominators,
00:40:27.500 --> 00:40:31.700
setting corresponding things equal, that you want to do by hand.
00:40:31.700 --> 00:40:39.200
That is it, that is partial fraction decomposition of a rational function,
00:40:39.200 --> 00:40:45.400
where the denominator can be factored into three distinct linear factors.
00:40:45.400 --> 00:40:49.000
There is no repeats, you do not have x².
00:40:49.000 --> 00:40:51.700
You do not have 3x – 1³.
00:40:51.700 --> 00:40:55.000
Multiplicity is 1 on each.
00:40:55.000 --> 00:40:59.500
Let us talk about case 2.
00:40:59.500 --> 00:41:10.700
Case 2 is when our rational function, this time, the denominator,
00:41:10.700 --> 00:41:19.000
when we factor the denominator, it ends up being a product.
00:41:19.000 --> 00:41:26.500
The factoring is a product of linear factors.
00:41:26.500 --> 00:41:32.500
Again, we are sticking with linear factors, some repeated.
00:41:32.500 --> 00:41:40.200
This time we have multiplicities which are going to be possibly greater than 1.
00:41:40.200 --> 00:41:54.900
That is, the denominator is going to end up looking like, it is going to be some a1x + b1 raised to some power,
00:41:54.900 --> 00:42:07.700
a2x + b2 raised to some power, a sub nx + bn raised to some power.
00:42:07.700 --> 00:42:23.600
The partial fraction decomposition looks like this.
00:42:23.600 --> 00:42:31.300
It looks like this, our nx/dx which is our rational function is equal to some constant,
00:42:31.300 --> 00:42:47.700
a1/ the factor a1x + b1¹ + a2/ the same factor raised to the next higher power.
00:42:47.700 --> 00:43:10.700
a1x + b1², and you keep going until you reach the nth power.
00:43:10.700 --> 00:43:14.500
An nth power, that is just for the first factor.
00:43:14.500 --> 00:43:32.100
And then, it is + b1/ the second, a2x + b2¹ + b2/ a 2x + b2²
00:43:32.100 --> 00:43:59.800
+ a sub n a2x + b2 ⁺nth, + c1/ however many factors you have.
00:43:59.800 --> 00:44:06.000
Each factor that you have, you are going to have that many terms all the way up to the nth power.
00:44:06.000 --> 00:44:18.900
anx + bn, all of this will make sense when you see a problem, it is very simple, to the first power + c2/ a sub nx + b sub n.
00:44:18.900 --> 00:44:22.200
Writing this out is exhausting.
00:44:22.200 --> 00:44:39.800
+… + c ⁺p/ a sub nx + b sub n ⁺p.
00:44:39.800 --> 00:44:44.900
Let us fluke with an example, I think that is the best way to make sense of this.
00:44:44.900 --> 00:44:52.900
We have got ax + 14/ x + 5² 2x – 2x – 2.
00:44:52.900 --> 00:44:56.300
Notice a couple of things, this one, the denominator is already factored for us.
00:44:56.300 --> 00:45:01.400
We do not have to do the factoring.
00:45:01.400 --> 00:45:06.200
The denominator is already factored, very nice.
00:45:06.200 --> 00:45:09.500
You will notice that the factors are linear.
00:45:09.500 --> 00:45:11.900
Exponent is 1, exponent is 1, they are linear.
00:45:11.900 --> 00:45:21.100
One of the factors is repeated twice.
00:45:21.100 --> 00:45:32.200
The x + 5 factor is repeated twice.
00:45:32.200 --> 00:45:36.900
Therefore, the partial fraction decomposition for that factor is going to have two terms.
00:45:36.900 --> 00:45:40.500
The partial fraction decomposition for this is going to have one term.
00:45:40.500 --> 00:45:42.400
We are going to have a total of three terms.
00:45:42.400 --> 00:45:44.200
Here is what it looks like.
00:45:44.200 --> 00:45:57.100
It is going to be 8x + 14/ x + 5² × 2x - 2 is equal to, we will take the first factor.
00:45:57.100 --> 00:45:58.700
We will deal with x + 5 first.
00:45:58.700 --> 00:46:09.400
It is going to be a/ x + 5¹ + b/ x + 5².
00:46:09.400 --> 00:46:11.800
Because that is 2 and that is 2, I can stop.
00:46:11.800 --> 00:46:19.300
Now I move to the next factor, + c/ 2x – 2¹.
00:46:19.300 --> 00:46:24.000
That is it, it is the first power that only shows up once.
00:46:24.000 --> 00:46:32.100
Now we do what we do.
00:46:32.100 --> 00:46:37.200
The least common denominator, I already know what that is.
00:46:37.200 --> 00:46:56.300
The least common denominator is x + 5² × 2x - 2 which actually is that thing.
00:46:56.300 --> 00:47:12.400
The numerator on the right becomes a × x + 5, because here x + 5 is only once.
00:47:12.400 --> 00:47:17.200
The least common denominator has it twice.
00:47:17.200 --> 00:47:39.200
a × x + 5 + b × 2x - 2 + c × x + 5² = ax + 5a.
00:47:39.200 --> 00:47:47.600
I’m sorry, I forgot one.
00:47:47.600 --> 00:47:52.000
It is a × x + 5 × 2x – 2.
00:47:52.000 --> 00:48:03.100
I forgot one, + b × 2x - 2 + c × x + 5².
00:48:03.100 --> 00:48:11.500
There we go, that equals a × 2x²,
00:48:11.500 --> 00:48:15.300
You can see how things can go soft very quickly.
00:48:15.300 --> 00:48:36.200
2x² + 8x - 10 + 2bx - 2b + c × x² + 10x + 25
00:48:36.200 --> 00:49:01.800
= 2ax² + 8ax - 10a + 2bx - 2b + cx² + 10cx + 25c.
00:49:01.800 --> 00:49:03.900
Let us go ahead and combine.
00:49:03.900 --> 00:49:14.300
x², x², we are going to take the x term, x term, x term.
00:49:14.300 --> 00:49:19.400
And then, number, number, number.
00:49:19.400 --> 00:49:22.100
Let us go ahead and do that.
00:49:22.100 --> 00:49:43.700
Our final is going to be 8x + 14/ x + 5² × 2x – 2, that is going to equal x² × 2a + c + x ×,
00:49:43.700 --> 00:50:09.000
when I combine those terms, 8a + 2b + 10c + -10a - 2b + 25c/ x + 5² × 2x – 2.
00:50:09.000 --> 00:50:12.600
Because the denominators are the same, the numerators are the same,
00:50:12.600 --> 00:50:21.200
and the only way they can be the same is if coefficients of corresponding terms are equal.
00:50:21.200 --> 00:50:35.200
Therefore, 2a + c = over here there is no x² term which means it is 0.
00:50:35.200 --> 00:50:41.800
8a + 2b + 10c, that is the coefficient of the x term.
00:50:41.800 --> 00:50:56.500
The coefficient of the x term is equal to 8 - 10a - 2b + 25c is equal to 14.
00:50:56.500 --> 00:51:00.100
There you go, this is the system of equations that you have to solve.
00:51:00.100 --> 00:51:03.600
Now I'm not going to go through the process of solving, I hope you will forgive me.
00:51:03.600 --> 00:51:07.500
I would like you to corroborate if you can, either by using software or doing it by hand.
00:51:07.500 --> 00:51:12.800
I have done it here but I really do not want to write everything else.
00:51:12.800 --> 00:51:19.700
I end up with a is equal to -11/36.
00:51:19.700 --> 00:51:35.200
I get b is equal to 167/72.
00:51:35.200 --> 00:51:40.100
I get c is equal to 11/18.
00:51:40.100 --> 00:51:43.400
Those are my three coefficients.
00:51:43.400 --> 00:52:01.600
Therefore, let us rewrite, we had 8x + 14 was our original rational function.
00:52:01.600 --> 00:52:06.700
We have x + 5² × 2x – 2.
00:52:06.700 --> 00:52:20.400
We said that it equals a/ x + 5 + b/ x + 5² + c/ 2x - 2
00:52:20.400 --> 00:52:46.300
that = -11/36 / x + 5 + 167/72 / x + 5² + 11/18 / 2x – 2.
00:52:46.300 --> 00:52:52.600
The integral of this is the integral of this.
00:52:52.600 --> 00:52:55.300
The integral of that breaks up into three integrals.
00:52:55.300 --> 00:53:21.100
Let me go back to blue, = -11/36 × the integral of 1/ x + 5 dx + 167/72 ×
00:53:21.100 --> 00:53:34.800
the integral of 1/ x + 5² dx + 11/18 × the integral of 1/ 2x - 2 dx.
00:53:34.800 --> 00:53:58.100
The answers I get are – 11/36 × the natlog of x + 5, an absolute value, + 167/72 × 1/ -1 × 1/ x + 5.
00:53:58.100 --> 00:54:05.600
I hope this integration does not throw you guys off.
00:54:05.600 --> 00:54:16.900
+ 11/18, this one is just set u equal to x + 5, do a u substitution.
00:54:16.900 --> 00:54:24.100
It ends up being the integral of u⁻² du, and then, some factor.
00:54:24.100 --> 00:54:29.900
11/18 × ½, the natlog of 2x – 2.
00:54:29.900 --> 00:54:34.400
This ½ term comes from the fact that that is a 2 + c.
00:54:34.400 --> 00:54:38.600
There you go, very nice.
00:54:38.600 --> 00:54:42.700
Very tedious but, beautiful process.
00:54:42.700 --> 00:54:44.800
We have taken care of two cases here.
00:54:44.800 --> 00:54:51.200
We have taken care of the case where the denominator factors into linear factors that are distinct.
00:54:51.200 --> 00:54:55.500
We also did the linear factors that possibly some are repeated.
00:54:55.500 --> 00:54:56.600
We are going to stop here.
00:54:56.600 --> 00:55:02.500
In the next lesson, we are going to do where it factors into a linear and quadratic factors.
00:55:02.500 --> 00:55:08.500
And then, linear and quadratic factors with some of the quadratic factors are now repeated.
00:55:08.500 --> 00:55:10.600
With that, thank you so much for joining us here at www.educator.com.
00:55:10.600 --> 00:55:12.000
We will see you next time, bye.