Original subtitles

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Today we're going to talk about how you use partial fractions to integrate a rational function with

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distinct quadratic factors in its denominator to complete this problem will set up the decomposition

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and solve for constants and then integrate the decomposition.

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In this particular problem we've been asked to use partial fractions to take the integral of the rational

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function that I've written here.

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Our first step with any partial fractions problem is to ensure that the degree of the denominator is

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greater than the degree of the numerator meaning that the largest exponent in the denominator is greater

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than the largest exponent in the numerator.

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In this case that's true so we can move on to the next step.

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If it weren't true for example if we add X to the fourth in the numerator and x cubed in the denominator

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we need to use polynomial long division to simplify our rational function.

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But in this case we can go ahead and move onto the next step and our next step will be to factor the

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denominator of our rational function as completely as possible.

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So the denominator will factor into the quantity x squared plus four times the quantity x squared plus

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1.

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We always want to factor as completely as possible in this case.

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This is as far as we can go.

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Once we've done map we want to set this equal to our partial fractions the composition which will put

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over here on the right hand side.

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So a partial fraction decomposition.

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What we do is we separate each term in the denominator.

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Each factor in the denominator into its own fraction.

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Over here on the right hand side.

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So we'll put the first factor X squared plus four into its own fraction and we'll put the second factor

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X squared plus 1 into its own fraction like this.

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Now the question is what do we need to put in the numerator of each of these fractions Well when we're

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dealing with quadratic factors meaning that the degree of the factor is 2 or greater.

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So we have x squared plus for the degree is 2 or greater.

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That means that it's a quadratic factor.

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If it's X to the first power or just X..

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So for example x plus four that's a linear factor and the process is different.

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But in this case we have two quadratic vectors and whenever we have a quadratic factor what we want

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to put in the numerator is a constant times x plus another constant.

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So in this case we'll say X plus B since this fraction over here has a quadratic factor in the denominator.

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We wanted to do the same thing.

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We've already used constants A and B.

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So in this fraction we'll put C X plus D to distinguish the difference between these constants.

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Now that we have the set up for our decomposition we want to go ahead and multiply both sides of this

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equation by the denominator from the left hand side.

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So we're going to multiply both the left and right hand side by the quantity x square plus four times

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the quantity x x squared plus 1.

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When we do that the entire denominator will cancel from the left hand side.

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And of course we'll just be left with x cubed minus 2 x squared plus X plus 1.

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When we multiply that denominator by the right hand side specifically this first fraction here the x

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squared plus four factor will cancel from the denominator here and we'll just be left with the quantity

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x plus being times the quantity x squared plus 1.

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The trick I would use for remembering this when I was taking tests is just I'm always left with a factor

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that's not in the denominator.

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So because X squared plus fours in the denominator here obviously I know that the other factor X squared

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plus 1 is the one that's going to be multiplied here by X plus B.

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Same thing over here when I multiply this entire denominator by this fraction.

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The x squared plus 1 will cancel and I'll be left with the quantity C X plus three times the quantity.

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The other factor X squared plus four.

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Now that I've eliminated all of my fractions what I want to go ahead and do is expand the right hand

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side over here.

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In other words multiply all these terms together.

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So when I multiply the quantity x plus B times the quantity x squared plus 1 I'll get a x cubed plus

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X plus b x squared plus B.

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And when I multiply the quantity C X plus the times the quantity x square plus 4 I'll get c x cubed

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plus for C X plus D x squared plus for me my goal now here is going to be to collect like terms so I

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want to pull together all of my terms involving x cubed all of my terms involving x squared all of my

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terms involving X and all of my constants.

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So notice here that I have a x cubed and I have C x cubed.

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So when I want to do is pull those two terms together and factor out the x cubed.

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When I do that I'll get a plus C times x cubed.

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I'm going to do the same thing for each of these other factors.

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So notice here that I have the x squared and DX squared.

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If I pull those together and factor out an x squared I'm going to get plus the quantity B plus D times

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x squared.

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When I bring mine a x and my 4C x terms together and I factor out an x I'll get plus the quantity a

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plus for C times x.

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And then when I bring my constants together b and for Dean I don't have to factor out anything because

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these are just constant.

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So I'll say plus the quantity B plus for D.

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The reason for collecting terms like this and factoring out the X variables is now I can equate coefficients

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on the left hand on the left and right hand side.

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So I can say the coefficient here on x cubed.

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Right now this is an empty box because the coefficient is just one.

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But I can say that the coefficient on the x cubed term here 1 is equal to the coefficient on the x cubed

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term on the right.

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A plus.

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See.

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So I can set those equal to each other and say one is equal to a plus C in the same way.

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I can set the convention on the left hand side on the x squared term negative two equal to the coefficient

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on the x squared term on the right hand side which is the posti.

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So I can say negative 2 is equal to B plus D.

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Again I can take the coefficient on the x term here which is just one and set that equal to a plus for

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C the coefficient on the x term on the right hand side.

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So say one is equal to a plus four scene.

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And then finally the constant on the left hand side one can be set equal to the constant on the right

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hand side over here B plus four.

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So I can say one is equal to B plus four D.

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This is basically now a system of simultaneous equations that I can use to solve for the constants A

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B C and D.

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I've collected my equations in this way.

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I've grouped the equations that have only a and c involved here on the left and the equations that only

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have B and D involved.

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Over here on the right that makes it easy for me to solve for the constant.

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So what I'll do over here on the left is subtract this entire first second equation here from the top

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equation like this.

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When I do that I'll get one minus one which will give me 0 on the left hand side.

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I'll get a minus a which again will give me 0 and I'll get C minus for C which will give me a negative

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3 C that gives me 0 equals negative 3 C which gives me C is equal to zero.

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When I plug that back in to my equation at the top here I'll get one equals a plus zero.

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And I can see that one is equal to a.

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So now I can start an documenting that I have solved for some of my constants.

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I'm going to do the same thing over here to solve for B and D.

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I'll subtract this entire second equation from the first equation.

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The reason I can do it is because I know looking at this equation that if I subtract the second one

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from the first one then my b's will cancel and I'm looking for one of the variables to disappear so

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I can solve the other one.

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I did the same thing over here with the A's and the C's.

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I knew that the A's would cancel when I said a minus a.

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So I didn't have to manipulate or do anything extra to this so I subtract that negative to minus 1 which

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should give me a negative 3 B minus B will just give me a 0 and D minus 4 D will give me a negative

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3 D when I divide both sides by negative 3.

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I can see that D is equal to 1.

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And when I plug the equals one back into this first equation over here I get negative two equals B plus

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1.

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Subtract one from both sides I get negative three equals B.

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So I can indicate that I've now also solved for B and D.

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What this means now that I have all four constants is that I can plug each of these four constants back

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into my partial fractions decomposition here and then integrate that.

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So I saved our constants up here in the upper right hand corner.

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What I'm going to do now is plug them back into our decomposition.

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So I'm going to have the integral of here a x plus B over the quantity x where plus 4.

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Well we know that a is 1.

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So we'll get 1 x and we know that B is negative 3.

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So we get plus negative 3 divided by the quantity x square plus 4.

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And then we have plus C X plus D when we know that C is zero.

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So we get 0 x plus D is 1 and will divide that either quantity x squared plus 1 D X..

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Now I want to simplify this so I'll get X minus 3 over x squared plus four plus one over the quantity

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x squared plus 1 D X with any partial fractions.

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The composition problem once you get to this point you're probably going to want to look at breaking

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these fractions apart as much as possible into separate integrals to try to make this easier.

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So what we'll do is we'll break apart this first fraction X minus three divided by x squared plus four

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into two and we'll break them into separate integrals so we'll get the integral of x squared plus for

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the X minus three times the integral of one over x squared plus for D x.

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So what we did there is we separated it into two fractions.

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This first term in the numerator X divided by x squared plus four.

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And the second term in the numerator negative three divided by x squared plus four.

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And in the process we pulled the negative three out in front of the integral to leave just one over

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x squared plus four.

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We can't simplifier a second fraction at all so we'll still just say plus one over the quantity x squared

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plus 1 the X and now that we have this simplified we want to go ahead and start integrating the first

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fraction here X divided by x squared plus four.

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We're going to need use substitution for will set you equal to x squared plus four.

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We'll take the derivative to get D-New which will be 2 x times d x and then we'll solve for the X to

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get the x is equal to d u over to X.

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Our second and third integrals will be using this formula up here that I didn't mention yet but the

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formula is for the integral of one over x squared plus any constant.

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So in this case with our second integral here a will be two.

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Because notice we're taking a constant Square is a constant and we're saying a squared.

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So we're looking for essentially the square root of the second term here.

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And the square root of four is two.

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So we know that a in this formula will be 2.

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And what this formula tells us is that 1 divided by x squared plus a constant squared is equal to 1

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divided by that constant times tangent to the negative one.

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Or the inverse tangent of X divided by that constant again.

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And of course C the constant of integration.

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So we're going to be using that formula for the second and third integrals that we have here.

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So let's go ahead and for this first integral make our substitutions with you.

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So we'll say X over you because remember we said X equal to the denominator x square plus 4.

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And then we solved for the X so for dx we got you over to X..

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So those two are multiplied together and then we have our second and third integrals here.

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So we're minus three.

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Now the integral of 1 divided by x squared plus four.

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Again we use this formula and notice that we're setting everything equal to this here so we'll get 1

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divided by a.

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We know that a is going to be 2 because we're taking the square root of this.

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The constants in this position here and square root here of four is two.

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So we'll get two times tangent to the negative one of x divided by again a as two we'll leave the constant

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immigration for the end of our integration.

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So now we take the integral of the third integral here and we'll get 1 divided by 1.

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Because the constant here is just one the square root of which is 1 so 1 over 1 tangent to the negative

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one of x divided by 1.

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And now we can add our plus c our constant integration.

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We want to go ahead and simplify as much as we can.

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Notice that we'll get the X here to cancel and we can pull the one half out in front so we'll get 1

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1/2 times the integral of one over you do you minus we'll say three halves tangent to the negative one

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of X over two plus tangent to the negative one of x plus C.

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And when we integrate one over you remember that the integral of one to over a variable like this is

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the natural log of the absolute value of that variable.

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So we'll get 1 1/2 times the natural log of the absolute value of you.

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And then the rest of our integration

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our final step is just to plug back in for you.

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So we'll get one halftimes the natural log of the absolute value.

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Remember that U is x squared plus 4.

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So x squared plus four and then minus three halves tangent to the negative one of x over two plus tangent

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the negative one of x and you can go ahead and leave it in this form sometimes you'll factor out of

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one half just to make this a little bit simpler and get rid of the coefficient I the natural log in

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the Tench the negative one here.

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But you don't need to.

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You can leave it in this form and this is your final answer.