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 continue our discussion of integration by partial fractions.
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Let us jump right on in.
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If you remember in the last lesson, we did the first two of the four cases,
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where once we factored the denominator of the rational expression as much as we can,
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if we have all linear factors or if some of those linear factors are repeated.
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Those are the first two cases that we dealt with last time.
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Now in this lesson, we are going to talk about quadratic factors and repeated quadratic factors, the last two cases.
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Let us start with case 3, I think I’m going to go ahead and work in blue.
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Case 3 is when our rational function, some m(x)/d(x),
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Again, this is just numerator and denominator as functions of x.
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That is when the d(x) contains irreducible quadratic factors.
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After you reduce, after you factored it as much as you can, one of the factors is quadratic.
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In other words, the highest degree on the x, the highest exponent is a 2,
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contains irreducible factors none of which is repeated.
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The first one is going to be none repeating irreducible quadratic factors is repeated.
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Let us recall irreducible means we cannot factor it any further.
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We cannot factor the quadratic any further into linear factors.
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That is our hope, we want to be able to factor something, any polynomial into, as many linear factors as possible.
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As it turns out, ultimately, you can only go to linear and quadratic.
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You can always you do that, but sometimes you cannot always factor the quadratic.
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That is the irreducible.
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Here is an example of irreducible is x² + 4.
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This one, you cannot reduce any further.
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Another example might be something like x² + 3x + 1.
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You cannot factor it any further.
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We said in the previous lesson that each linear factor ax + b gives a partial fraction of a/ ax + b.
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If one of the factors is linear on top, the variable, the thing that we are looking for is just a constant, it is just a.
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Now for quadratic factors, each factor of the form ax² + bx + c, because the quadratic factor is going to be some variation of this.
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Sometimes the bx term would not be there.
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Sometimes the c would not be there.
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It is going to look something like that.
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It gives ax + b/ ax² + bx + c.
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On the top, our unknown, where it is actually going to be a full linear factor.
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The thing to notice here is the denominators of degree 1, the numerator is 1° less.
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That is why it is just a, it is x⁰.
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Here the denominator, the highest degree is 2.
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Therefore, on the numerator, it is going to be 1° less which means this type of function ax + b, a linear factor.
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For linear factors, we put a on top, a constant.
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For quadratic factors, we put ax + b.
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Sorry, we use capitals, ax + b.
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Where a, b, and c, and d, it could be, if you have another quadratic factor, it would be cx + d, ex + f, and so on.
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Let us go ahead and do an example, I think it will make sense.
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We want to evaluate the integral 12/ x - 2 × x² + 9.
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The denominator here, again, the first thing we do is factor the denominator as much as we can.
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Here the denominators are already factored.
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x - 2 is our linear factor, x² + 9 is our irreducible quadratic factor.
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This does not factor anymore.
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If it were x² – 9, that is fine, we can do x + 3x - 3 but were stuck like this.
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The denominators are already factored.
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Now 12, this 12/ x - 2 × x² + 9, it is going to equal, we have a linear factor x – 2.
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The partial fraction decomposition is going to be a/ x - 2 + this is our quadratic factor.
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This is going to be x² + 9 and we put bx + c.
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Our task is to find the a, find the b, and find the c, so that we have a partial fraction decomposition of our original rational function.
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Once we separate that, we are going to integrate each one separately.
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Let us go ahead and do that.
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This is going to be, I actually solve for the least common denominator here.
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This is going to be a × x² + 9 + bx + c × x - 2/ the least common denominator which is x - 2 × x² + 9.
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We are going to concern ourselves only with the numerator.
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Because now we have this equal to this, the denominators, this and this are the same,
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that means that the numerator and the numerator are the same.
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I'm just going to work with the numerators.
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Once you actually do that on the right, once you find a common denominator, put it under a common denominator.
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The denominators go away.
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They are equal, therefore, the numerators are equal.
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We concern ourselves only with the numerators.
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Therefore, we have 12 is equal to, I can multiply this all this out here.
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We have ax² + 9a + bx² + cx - 2bx – 2c.
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12 =, I’m going to take care of the ax² bx².
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It is going to be x² × a + b, that takes care of the ax² + bx².
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We will do the x terms.
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The x terms, I have c - 2b that takes care of the x terms.
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I have + 9a - 2c, 9a - 2c that takes care of the number terms.
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Now I equate coefficients.
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Over on the left, there is no x² term.
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Therefore, a + b is equal to 0, 0x².
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Therefore, I have the equation, a + b is equal to 0.
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On the left, there is no x term, therefore, it is 0x.
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Therefore, c - 2b is equal to 0.
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The number 12 is that one.
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Therefore, I have 9a - 2c is equal to 12.
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These three, these three equations and three unknowns is what we are going to solve for a, b, and c.
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Let us go ahead and do that next.
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I have got a + b = 0, I’m going to write them this way.
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C - 2b = 0 and 9a - 2c = 12.
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I presume that most of you are comfortable with solving two and three, sometimes four equations, and that many unknowns.
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But I will go through the process anyway, it is not a problem.
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It only takes a couple of minutes.
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Here a = -b, here c = 2b.
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I’m going to put this a and this c into here and solve for b, and then put the b’s back and find a and c.
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I have 9 × a which is - b - 2 × 2b which is c equal to 12.
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I have -9b - 4b = 12.
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Sorry, this looks like a 13, this is a b.
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I have got -13b is equal to 12.
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Therefore, I find that b is equal to -12/13.
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That takes care of b.
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Now I go ahead and put that over here.
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I find that a = - a -12/13 which means that a is equal to 12/13.
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Of course, c is equal of 2 × b which is -12/13.
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Therefore, c is equal to -24/13.
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Now I found a, b, and c, I put them back into my original decomposition.
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Remember, we had that our original 12/ x - 2 × x² + 9 is equal to, our decomposition was a/ x - 2 + bx + c/ x² + 9.
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Therefore, we just put it in.
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I have a is 12/13, this is going to be 12/13 / x - 2 + b which is -12/13 x + c which is -24/13 / x² + 9.
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There you go, this is our partial fraction decomposition.
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This is the first part, we did our partial fraction decomposition.
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Now we actually want to integrate this.
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The integral of this is going to be the integral of this.
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The integral of this is going to equal the integral of this + the integral of that.
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That is it, just work your way through.
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Let us see what we have got here.
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Our integral, which I will just call int, is equal to the integral of 12/13 / x - 2 dx
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+ the integral of -12/13 x - 24/13 / x² + 9 dx, which is going to end up equaling 12/13
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× the integral of dx/ x - 2 - 12/13 × the integral of x/ x² + 9.
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Just separating this thing out -24/13 the integral of, this is x dx, sorry about that.
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I always forget the dx, for all these years, I still forget it.
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x² + 9.
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We end up getting the following.
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We end up getting 12/13 × the natlog of the absolute value of x - 2 - 12/13 × ½ the natlog of x² + 9.
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This ½ factor came from the fact that this is a u substitution.
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I let u equal x², therefore, du = 2x dx x dx.
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Bring the 2, u substitution, I will let you work that out.
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This one is going to be -24/13 √9.
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I will tell you where this came from in just a minute.
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1x/ √9 + c, there we go.
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The last integral, this one right here, where did I get that?
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Here is where I got that.
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For the last integral, we use the following formula.
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The integral of 1/ x² + a² dx is equal to 1/a × tan⁻¹(x)/ a.
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That is the form that we use for this because often, when we do partial fraction decompositions,
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especially when we have quadratic factors in the denominator, we often end up with integrals that look like that.
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Some dx which is just the 1 dx over here/ something x² + something else x² + something else².
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It is the general integral that keep showing up.
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We went ahead and we are just going to use the formula for it.
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1/a × tan⁻¹(x)/ a.
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Let us do another example here, this time we want to evaluate the integral of x – 4/ 4x² + 4x +5.
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We take a look at this and we realize that the denominator cannot be factored any further.
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This is definitely an irreducible quadratic factor.
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4x² - 4x + 5 is irreducible and it is also the only factor.
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Because it is the only factor, there is no partial fraction decomposition.
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The rational function itself, it is the partial fraction decomposition.
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It is a partial fraction decomposition that is composed of just one term, x - 4/ 4x² – 4x + 5.
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There are no other factors in the denominator, for me to actually expand and do what I have done in the previous problems.
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Here we are going to show you a general procedure for how to handle a quadratic in the denominator that is the only term.
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Here is how we do it.
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We actually are going to complete the square in the denominator.
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We handle the situation, again, this is a general procedure,
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anytime you have a quadratic in the denominator that it is the only factor.
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It is a single only factor.
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We handle this situation by completing the square.
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Something that you have done thousands of times in algebra, by completing the square in the denominator.
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I got to tell you the technique of completing the square is something that comes in handy so often,
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and so many other branches of mathematics.
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We are working just with the denominator.
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The denominator, we got 4x² - 4x + 5.
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I’m going to go ahead and factor out the 4.
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This is going to be 4 × x² – x.
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I will leave a little space for something that I add, + 5.
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I take half of the second term which is -1/2 and I square it.
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This is going to be + ¼.
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Since I added 4 × 1/4, I added 1, I'm going to subtract 1 from this expression to retain the equality.
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I'm going to write this as 4 × x – ½² + 4.
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This is just 2² × x – ½² + 4.
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Again, this is just mathematical manipulation, nothing strange happening here.
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This is 2² × something², I’m going to put them together and take the squared out.
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This is going to be 2 × x - ½² + 4.
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I'm going to multiply, I’m going to distribute the 2 in there.
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This is going to end up being 2x - 1² + 4.
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Now I have that, this is my denominator.
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I just changed the way it looks.
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We have the integral of x - 4/ 2x - 1² + 4 dx.
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Now I’m going to subject this to a u substitution.
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Let us do this in red.
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I’m going to let u equal 2x – 1.
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I’m going to let du = 2 dx, that means dx is equal to du/2.
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I have taken care of that.
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Over here I’m going to actually solve for x.
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It is going to be x is going to equal, because I want to also deal with this x on top, in terms of u.
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x, when I solve this equation, we said u = 2x - 1 so x = u + 1/ 2.
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When I put all of these in here, we have the integral of u + 1.
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I’m going to write it as ½ of u + 1, that is my x, - 4/ u² + 4 dx is du/2 which = ½.
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I’m going to pull this ½ out.
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The integral of ½ u + ½ - 4 which I’m going to write as 8/2 / u² + 4 du
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= ½ of the integral of ½ u - 7/2 / u² + 4 du.
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I’m going to pull the ½ here, put it here, it equals ½ × ½ the integral of u - 7/ u² + 4.
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So far so good, that = ¼ × the integral of u/ u² + 4 du - 7/4 × the integral of 1/ u² + 4 du.
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Now these integrals, I can handle.
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Remember, again, u = 2x – 1.
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Let us actually write this again.
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We said that it equal ¼ × the integral of u/ u² + 4 du – 7/4 × the integral of 1/ u² + 4 du.
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That gives us ¼ × ½ × natlog of u² + u – 7/4 × ½ tan⁻¹ of u/ 2 + c.
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This ½ term, that one comes from the fact that I do a second u substitution on this.
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If I call that one v, v = u².
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This ½ comes from the fact that this is that formula, it is 1/a u² + 2².
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Remember that formula that we just did.
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We said that the integral of 1/ u² + a² = 1/a × tan⁻¹ of 1/a.
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Here this is a² which means that a is 2.
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That is where this one actually comes from.
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Sorry this looks like a u, a² + 4, this is the u and this is the 4.
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There you go, again, this is a general procedure that you can use whenever you have an integral where you have only one factor.
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Or the rational functional only has one factor and it is an irreducible quadratic.
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You complete the square on that thing and then you use a u substitution to do what we just did.
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It will always work.
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Let me actually write that down.
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This problem offers a general procedure for dealing with integrals of the type,
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on the very attractive integral sign, it is b/ ax² + bx + c dx.
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Anytime you are faced with an integral that looks like that, you can run this procedure.
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Complete the square and then solve the integral.
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Now let us deal with our 4th and final case.
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We just did irreducible quadratic factors that are non-repeating.
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What if we have repeating quadratic factors?
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That is actually going to be the same thing.
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It is just more terms in your partial fraction decomposition.
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Case 4, it is where our numerator/ our denominator is our rational function.
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It is where our denominator has repeated irreducible quadratic factors.
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In other words, it is a quadratic factor of the form ax² + bx + c raised to some power.
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It itself might be squared, a quadratic factor might show up two times, three times, four times.
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It is called the algebraic multiplicity, the multiplicity of the factor.
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The partial fraction decomposition of something like this is as follows.
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You have a1 x + b1/ ax² + bx + c + a2 x + b2/ ax² + bx + c²,
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and so on, until you get to a sub n x + b sub n/ ax² + bx + c ⁺n.
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In other words, whatever n is, you are going to have that many.
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You are going to have that first power, second power, all the way up to nth power.
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You are going to have all of these linear factors up on top.
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You have all of these coefficients to find.
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Let us do an example, I think it will make sense.
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Same exact thing that what we did for the repeated linear factors, just have them keep showing up to the nth power.
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Evaluate 2/ x × x² + 6².
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In this case, this quadratic factor is irreducible, x² + 6 cannot be factored.
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It itself is raised to the second power.
00:30:32.900 --> 00:30:46.000
The denominator is already factored, we do not have to do that.
00:30:46.000 --> 00:30:55.000
It is already factored, therefore, 2 divided by x, x² + 6².
00:30:55.000 --> 00:31:00.000
We have a linear factor that is the x, it becomes a/x.
00:31:00.000 --> 00:31:02.600
We have a quadratic factor raised to the power of 2.
00:31:02.600 --> 00:31:19.000
We are going to do bx + c/ this quadratic factor to the first power + dx + e, the quadratic factor raised to the second power.
00:31:19.000 --> 00:31:25.800
Our partial fraction decomposition involves one term where the quadratic factor is to the first power
00:31:25.800 --> 00:31:29.700
and the second term where the quadratic factor is to the second power.
00:31:29.700 --> 00:31:31.800
As many terms, all the way up to that many powers.
00:31:31.800 --> 00:31:39.300
If this were a 3, we would have fx + g/ x² + 1³, and so on.
00:31:39.300 --> 00:31:49.200
This is it, now we are going to find the least common denominator on this side.
00:31:49.200 --> 00:31:51.900
Let me actually write this out.
00:31:51.900 --> 00:31:57.300
The least common denominator here, be very careful, the least common denominator is not this × this × this.
00:31:57.300 --> 00:32:05.900
It is this × this because this factor is already contained in that.
00:32:05.900 --> 00:32:19.900
Our lowest common denominator is x × x² + 6², and whatever is on top.
00:32:19.900 --> 00:32:25.600
Be very careful, you are used to having the least common denominator, just multiply the denominators.
00:32:25.600 --> 00:32:30.900
Here, because the factors are repeated, this does not need to be this × this × this.
00:32:30.900 --> 00:32:33.300
It does not need to be that way, this is already contained in that.
00:32:33.300 --> 00:32:35.500
It is just this and this.
00:32:35.500 --> 00:33:00.800
Therefore, here what we need to do is we need to do a × x² + 6² + bx + c × x² + 6, only once.
00:33:00.800 --> 00:33:23.200
And + dx + b × x/ x × x² + 6².
00:33:23.200 --> 00:33:29.500
This denominator is the same as that denominator, which means the numerator is equivalent to the numerator.
00:33:29.500 --> 00:33:36.000
Now we expand this numerator and that is what we are going to do next.
00:33:36.000 --> 00:33:45.300
2 is going to equal, when I multiply all this out.
00:33:45.300 --> 00:33:49.900
That is fine, it is just algebra.
00:33:49.900 --> 00:34:30.100
ax⁴ + 12ax² + 36a +, it is going to end up being bx⁴ + cx³ + 6bx² + 6cx + dx² + e ⁺x.
00:34:30.100 --> 00:34:37.300
When we combine terms, we have an x⁴ term.
00:34:37.300 --> 00:34:44.400
This is going to be a, it takes care of that one, and a b.
00:34:44.400 --> 00:34:47.400
It takes care of the x⁴ terms.
00:34:47.400 --> 00:34:49.500
There is an x³ term.
00:34:49.500 --> 00:34:58.800
The only x³ term is that one, c.
00:34:58.800 --> 00:35:13.200
There is an x² term, 12a + 6b + d.
00:35:13.200 --> 00:35:16.200
Make sure you get all the terms.
00:35:16.200 --> 00:35:31.300
There is an x term, 6c + e.
00:35:31.300 --> 00:35:38.200
There is a number term, 36a.
00:35:38.200 --> 00:35:42.700
All of that is equal to 2.
00:35:42.700 --> 00:35:46.300
We just set things equal to each other.
00:35:46.300 --> 00:35:52.800
The equations that we get, in other words, a + b is going to be 0, c is going to be 0.
00:35:52.800 --> 00:35:57.600
12a + 6b + d is going to be 0.
00:35:57.600 --> 00:36:00.400
6c + e is going to be 0.
00:36:00.400 --> 00:36:06.700
36a is going to equal 2, that is what we get.
00:36:06.700 --> 00:36:11.700
We are going to get a + b = 0.
00:36:11.700 --> 00:36:24.200
We are going to get c = 0, we are going to get 12a + 6b + d is equal to 0.
00:36:24.200 --> 00:36:29.300
We are going to get 6c + e is equal to 0.
00:36:29.300 --> 00:36:32.700
We are going to get 36a is equal to 2.
00:36:32.700 --> 00:36:37.800
This gives us that a is equal to 1/18.
00:36:37.800 --> 00:36:40.800
That takes care of that.
00:36:40.800 --> 00:36:43.200
Let us go to the first one over here.
00:36:43.200 --> 00:37:02.500
We use the equation a + b is equal to 0, which means that a = -b which means that b = -a which means of b = -1/18.
00:37:02.500 --> 00:37:07.300
We already know that c is equal to 0.
00:37:07.300 --> 00:37:09.400
We have taken care of that.
00:37:09.400 --> 00:37:17.200
Now we have 6c + e is equal to 0.
00:37:17.200 --> 00:37:27.400
c is equal to 0 so I get 0 + e is equal to 0, which means that e is also equal to 0.
00:37:27.400 --> 00:37:32.700
Now I’m going to use this equation.
00:37:32.700 --> 00:37:46.700
12 × 1/18 + 6b, 6 × -1/18 + d is equal to 0.
00:37:46.700 --> 00:37:53.300
When I solve this, I get d is equal to -6/18.
00:37:53.300 --> 00:37:56.500
I’m going to leave it as -6/18, instead of reducing that.
00:37:56.500 --> 00:38:09.100
Therefore, our final partial fraction decomposition, our original function was 2/ x × x² + 6².
00:38:09.100 --> 00:38:24.200
We said that was equal to a/x + bx + c/ x² + 6 + dx + e/ x² + 6².
00:38:24.200 --> 00:38:28.100
Now we have a, b, c, d, and e.
00:38:28.100 --> 00:38:55.800
Our decomposition is 1/18 / x + -1/18 x + c which is 0/ x² + 6 + -6/18 x + e which is 0/ x² + 6².
00:38:55.800 --> 00:39:00.000
This is our final partial fraction decomposition.
00:39:00.000 --> 00:39:02.400
That is just the decomposition, that is not the answer.
00:39:02.400 --> 00:39:06.000
We still have to integrate this thing.
00:39:06.000 --> 00:39:11.300
The integral of this is the integral of this because these are the same.
00:39:11.300 --> 00:39:17.100
The integral of this is the integral of this + the integral of that + the integral of that.
00:39:17.100 --> 00:39:22.200
Nice and simple, the rest is just using all the techniques that we gathered so far.
00:39:22.200 --> 00:39:38.300
My final answer, our integral is going to equal 1/18 × the integral of 1/x dx - 1/18
00:39:38.300 --> 00:40:03.600
× the integral of x/ x² + 6 dx – 6/18 × the integral of x/ x² + 6² dx.
00:40:03.600 --> 00:40:17.400
This is equal to 1/18 × natlog of the absolute value of x - 1/18 × ½, because of that u.
00:40:17.400 --> 00:40:43.700
u = x² + 6, du = 2x dx, x dx = du/2 × natlog of x² + 6 – 6/18 × the integral of x/ x² + 6² dx.
00:40:43.700 --> 00:40:48.500
How do we handle this?
00:40:48.500 --> 00:40:52.300
We handle it this way, we do we a u substitution on this.
00:40:52.300 --> 00:41:07.000
We let u = x² + 6, du = 2x dx, du/2 = x.
00:41:07.000 --> 00:41:10.200
It is essentially the same thing that we did here, because of that extra x² part,
00:41:10.200 --> 00:41:18.600
I thought it actually do the u substitution, x dx.
00:41:18.600 --> 00:41:48.900
We have ½ × the integral of u⁻² du which is equal to ½ × u⁻¹/ -1 which = -1/ 2u which is equal to -1/ 2 × x² + 6.
00:41:48.900 --> 00:42:10.900
Our final answer is = 1/18 × natlog of x – 1/18 × ½ × natlog of x² + 6
00:42:10.900 --> 00:42:27.000
– 6/18 × -1/ 2 × x² + 6 + c.
00:42:27.000 --> 00:42:32.400
There we go, partial fraction decompositions are very tedious.
00:42:32.400 --> 00:42:35.300
They are algebraically intense.
00:42:35.300 --> 00:42:40.500
There are plenty of places where you can make a mistake but conceptually I do not think it is all together that difficult.
00:42:40.500 --> 00:42:42.600
You just have to keep track of everything.
00:42:42.600 --> 00:42:48.900
But this is calculus, you are more than accustomed to that by now because the problems just are,
00:42:48.900 --> 00:42:54.000
by nature, sort of long and detailed.
00:42:54.000 --> 00:42:56.100
Thank you so much for joining us here at www.educator.com.
00:42:56.100 --> 00:42:57.000
We will see you next time, bye.