AP Calculus AB Integration by Partial Fractions II

Hello, welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to continue our discussion of integration by partial fractions.0004

Let us jump right on in.0008

If you remember in the last lesson, we did the first two of the four cases,0011

where once we factored the denominator of the rational expression as much as we can,0016

if we have all linear factors or if some of those linear factors are repeated.0023

Those are the first two cases that we dealt with last time.0028

Now in this lesson, we are going to talk about quadratic factors and repeated quadratic factors, the last two cases.0031

Let us start with case 3, I think I’m going to go ahead and work in blue.0039

Case 3 is when our rational function, some m(x)/d(x),0044

Again, this is just numerator and denominator as functions of x.0051

That is when the d(x) contains irreducible quadratic factors.0055

After you reduce, after you factored it as much as you can, one of the factors is quadratic.0064

In other words, the highest degree on the x, the highest exponent is a 2,0068

contains irreducible factors none of which is repeated.0074

The first one is going to be none repeating irreducible quadratic factors is repeated.0090

Let us recall irreducible means we cannot factor it any further.0103

We cannot factor the quadratic any further into linear factors.0115

That is our hope, we want to be able to factor something, any polynomial into, as many linear factors as possible.0134

As it turns out, ultimately, you can only go to linear and quadratic.0141

You can always you do that, but sometimes you cannot always factor the quadratic.0145

That is the irreducible.0149

Here is an example of irreducible is x² + 4.0151

This one, you cannot reduce any further.0161

Another example might be something like x² + 3x + 1.0164

You cannot factor it any further.0172

We said in the previous lesson that each linear factor ax + b gives a partial fraction of a/ ax + b.0177

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.0212

Now for quadratic factors, each factor of the form ax² + bx + c, because the quadratic factor is going to be some variation of this.0219

Sometimes the bx term would not be there.0248

Sometimes the c would not be there.0250

It is going to look something like that.0255

It gives ax + b/ ax² + bx + c.0257

On the top, our unknown, where it is actually going to be a full linear factor.0267

The thing to notice here is the denominators of degree 1, the numerator is 1° less.0273

That is why it is just a, it is x⁰.0278

Here the denominator, the highest degree is 2.0281

Therefore, on the numerator, it is going to be 1° less which means this type of function ax + b, a linear factor.0285

For linear factors, we put a on top, a constant.0295

For quadratic factors, we put ax + b.0298

Sorry, we use capitals, ax + b.0301

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.0304

Let us go ahead and do an example, I think it will make sense.0315

We want to evaluate the integral 12/ x - 2 × x² + 9.0321

The denominator here, again, the first thing we do is factor the denominator as much as we can.0326

Here the denominators are already factored.0331

x - 2 is our linear factor, x² + 9 is our irreducible quadratic factor.0333

This does not factor anymore.0338

If it were x² – 9, that is fine, we can do x + 3x - 3 but were stuck like this.0339

The denominators are already factored.0346

Now 12, this 12/ x - 2 × x² + 9, it is going to equal, we have a linear factor x – 2.0358

The partial fraction decomposition is going to be a/ x - 2 + this is our quadratic factor.0373

This is going to be x² + 9 and we put bx + c.0381

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.0386

Once we separate that, we are going to integrate each one separately.0395

Let us go ahead and do that.0398

This is going to be, I actually solve for the least common denominator here.0407

This is going to be a × x² + 9 + bx + c × x - 2/ the least common denominator which is x - 2 × x² + 9.0411

We are going to concern ourselves only with the numerator.0433

Because now we have this equal to this, the denominators, this and this are the same,0435

that means that the numerator and the numerator are the same.0447

I'm just going to work with the numerators.0451

Once you actually do that on the right, once you find a common denominator, put it under a common denominator.0454

The denominators go away.0460

They are equal, therefore, the numerators are equal.0461

We concern ourselves only with the numerators.0464

Therefore, we have 12 is equal to, I can multiply this all this out here.0476

We have ax² + 9a + bx² + cx - 2bx – 2c.0482

12 =, I’m going to take care of the ax² bx².0506

It is going to be x² × a + b, that takes care of the ax² + bx².0510

We will do the x terms.0518

The x terms, I have c - 2b that takes care of the x terms.0520

I have + 9a - 2c, 9a - 2c that takes care of the number terms.0528

Now I equate coefficients.0540

Over on the left, there is no x² term.0542

Therefore, a + b is equal to 0, 0x².0545

Therefore, I have the equation, a + b is equal to 0.0549

On the left, there is no x term, therefore, it is 0x.0553

Therefore, c - 2b is equal to 0.0557

The number 12 is that one.0561

Therefore, I have 9a - 2c is equal to 12.0565

These three, these three equations and three unknowns is what we are going to solve for a, b, and c.0571

Let us go ahead and do that next.0576

I have got a + b = 0, I’m going to write them this way.0582

C - 2b = 0 and 9a - 2c = 12.0586

I presume that most of you are comfortable with solving two and three, sometimes four equations, and that many unknowns.0596

But I will go through the process anyway, it is not a problem.0603

It only takes a couple of minutes.0605

Here a = -b, here c = 2b.0607

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.0614

I have 9 × a which is - b - 2 × 2b which is c equal to 12.0621

I have -9b - 4b = 12.0633

Sorry, this looks like a 13, this is a b.0641

I have got -13b is equal to 12.0644

Therefore, I find that b is equal to -12/13.0648

That takes care of b.0654

Now I go ahead and put that over here.0655

I find that a = - a -12/13 which means that a is equal to 12/13.0657

Of course, c is equal of 2 × b which is -12/13.0669

Therefore, c is equal to -24/13.0675

Now I found a, b, and c, I put them back into my original decomposition.0680

Remember, we had that our original 12/ x - 2 × x² + 9 is equal to, our decomposition was a/ x - 2 + bx + c/ x² + 9.0686

Therefore, we just put it in.0706

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 + 9.0707

There you go, this is our partial fraction decomposition.0735

This is the first part, we did our partial fraction decomposition.0742

Now we actually want to integrate this.0745

The integral of this is going to be the integral of this.0747

The integral of this is going to equal the integral of this + the integral of that.0751

That is it, just work your way through.0755

Let us see what we have got here.0759

Our integral, which I will just call int, is equal to the integral of 12/13 / x - 2 dx0766

+ the integral of -12/13 x - 24/13 / x² + 9 dx, which is going to end up equaling 12/130779

× the integral of dx/ x - 2 - 12/13 × the integral of x/ x² + 9.0809

Just separating this thing out -24/13 the integral of, this is x dx, sorry about that.0824

I always forget the dx, for all these years, I still forget it.0836

x² + 9.0841

We end up getting the following.0846

We end up getting 12/13 × the natlog of the absolute value of x - 2 - 12/13 × ½ the natlog of x² + 9.0847

This ½ factor came from the fact that this is a u substitution.0866

I let u equal x², therefore, du = 2x dx x dx.0870

Bring the 2, u substitution, I will let you work that out.0876

This one is going to be -24/13 √9.0882

I will tell you where this came from in just a minute.0890

1x/ √9 + c, there we go.0894

The last integral, this one right here, where did I get that?0907

Here is where I got that.0914

For the last integral, we use the following formula.0916

The integral of 1/ x² + a² dx is equal to 1/a × tan⁻¹(x)/ a.0927

That is the form that we use for this because often, when we do partial fraction decompositions,0943

especially when we have quadratic factors in the denominator, we often end up with integrals that look like that.0947

Some dx which is just the 1 dx over here/ something x² + something else x² + something else².0953

It is the general integral that keep showing up.0966

We went ahead and we are just going to use the formula for it.0970

1/a × tan⁻¹(x)/ a.0972

Let us do another example here, this time we want to evaluate the integral of x – 4/ 4x² + 4x +5.0981

We take a look at this and we realize that the denominator cannot be factored any further.0990

This is definitely an irreducible quadratic factor.1007

4x² - 4x + 5 is irreducible and it is also the only factor.1014

Because it is the only factor, there is no partial fraction decomposition.1039

The rational function itself, it is the partial fraction decomposition.1043

It is a partial fraction decomposition that is composed of just one term, x - 4/ 4x² – 4x + 5.1048

There are no other factors in the denominator, for me to actually expand and do what I have done in the previous problems.1055

Here we are going to show you a general procedure for how to handle a quadratic in the denominator that is the only term.1061

Here is how we do it.1070

We actually are going to complete the square in the denominator.1074

We handle the situation, again, this is a general procedure,1079

anytime you have a quadratic in the denominator that it is the only factor.1082

It is a single only factor.1086

We handle this situation by completing the square.1089

Something that you have done thousands of times in algebra, by completing the square in the denominator.1096

I got to tell you the technique of completing the square is something that comes in handy so often,1107

and so many other branches of mathematics.1114

We are working just with the denominator.1120

The denominator, we got 4x² - 4x + 5.1125

I’m going to go ahead and factor out the 4.1134

This is going to be 4 × x² – x.1136

I will leave a little space for something that I add, + 5.1142

I take half of the second term which is -1/2 and I square it.1146

This is going to be + ¼.1149

Since I added 4 × 1/4, I added 1, I'm going to subtract 1 from this expression to retain the equality.1156

I'm going to write this as 4 × x – ½² + 4.1164

This is just 2² × x – ½² + 4.1174

Again, this is just mathematical manipulation, nothing strange happening here.1181

This is 2² × something², I’m going to put them together and take the squared out.1185

This is going to be 2 × x - ½² + 4.1191

I'm going to multiply, I’m going to distribute the 2 in there.1202

This is going to end up being 2x - 1² + 4.1204

Now I have that, this is my denominator.1217

I just changed the way it looks.1226

We have the integral of x - 4/ 2x - 1² + 4 dx.1235

Now I’m going to subject this to a u substitution.1247

Let us do this in red.1254

I’m going to let u equal 2x – 1.1255

I’m going to let du = 2 dx, that means dx is equal to du/2.1263

I have taken care of that.1274

Over here I’m going to actually solve for x.1278

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.1280

x, when I solve this equation, we said u = 2x - 1 so x = u + 1/ 2.1291

When I put all of these in here, we have the integral of u + 1.1299

I’m going to write it as ½ of u + 1, that is my x, - 4/ u² + 4 dx is du/2 which = ½.1316

I’m going to pull this ½ out.1338

The integral of ½ u + ½ - 4 which I’m going to write as 8/2 / u² + 4 du1339

= ½ of the integral of ½ u - 7/2 / u² + 4 du.1363

I’m going to pull the ½ here, put it here, it equals ½ × ½ the integral of u - 7/ u² + 4.1374

So far so good, that = ¼ × the integral of u/ u² + 4 du - 7/4 × the integral of 1/ u² + 4 du.1389

Now these integrals, I can handle.1413

Remember, again, u = 2x – 1.1416

Let us actually write this again.1425

We said that it equal ¼ × the integral of u/ u² + 4 du – 7/4 × the integral of 1/ u² + 4 du.1437

That gives us ¼ × ½ × natlog of u² + u – 7/4 × ½ tan⁻¹ of u/ 2 + c.1455

This ½ term, that one comes from the fact that I do a second u substitution on this.1485

If I call that one v, v = u².1494

This ½ comes from the fact that this is that formula, it is 1/a u² + 2².1502

Remember that formula that we just did.1511

We said that the integral of 1/ u² + a² = 1/a × tan⁻¹ of 1/a.1515

Here this is a² which means that a is 2.1525

That is where this one actually comes from.1530

Sorry this looks like a u, a² + 4, this is the u and this is the 4.1540

There you go, again, this is a general procedure that you can use whenever you have an integral where you have only one factor.1550

Or the rational functional only has one factor and it is an irreducible quadratic.1560

You complete the square on that thing and then you use a u substitution to do what we just did.1564

It will always work.1571

Let me actually write that down.1577

This problem offers a general procedure for dealing with integrals of the type,1581

on the very attractive integral sign, it is b/ ax² + bx + c dx.1611

Anytime you are faced with an integral that looks like that, you can run this procedure.1627

Complete the square and then solve the integral.1631

Now let us deal with our 4th and final case.1637

We just did irreducible quadratic factors that are non-repeating.1640

What if we have repeating quadratic factors?1644

That is actually going to be the same thing.1646

It is just more terms in your partial fraction decomposition.1647

Case 4, it is where our numerator/ our denominator is our rational function.1654

It is where our denominator has repeated irreducible quadratic factors.1664

In other words, it is a quadratic factor of the form ax² + bx + c raised to some power.1686

It itself might be squared, a quadratic factor might show up two times, three times, four times.1713

It is called the algebraic multiplicity, the multiplicity of the factor.1719

The partial fraction decomposition of something like this is as follows.1740

You have a1 x + b1/ ax² + bx + c + a2 x + b2/ ax² + bx + c²,1743

and so on, until you get to a sub n x + b sub n/ ax² + bx + c ⁺n.1770

In other words, whatever n is, you are going to have that many.1783

You are going to have that first power, second power, all the way up to nth power.1789

You are going to have all of these linear factors up on top.1794

You have all of these coefficients to find.1799

Let us do an example, I think it will make sense.1805

Same exact thing that what we did for the repeated linear factors, just have them keep showing up to the nth power.1809

Evaluate 2/ x × x² + 6².1820

In this case, this quadratic factor is irreducible, x² + 6 cannot be factored.1824

It itself is raised to the second power.1829

The denominator is already factored, we do not have to do that.1833

It is already factored, therefore, 2 divided by x, x² + 6².1846

We have a linear factor that is the x, it becomes a/x.1855

We have a quadratic factor raised to the power of 2.1860

We are going to do bx + c/ this quadratic factor to the first power + dx + e, the quadratic factor raised to the second power.1862

Our partial fraction decomposition involves one term where the quadratic factor is to the first power1879

and the second term where the quadratic factor is to the second power.1886

As many terms, all the way up to that many powers.1890

If this were a 3, we would have fx + g/ x² + 1³, and so on.1892

This is it, now we are going to find the least common denominator on this side.1899

Let me actually write this out.1909

The least common denominator here, be very careful, the least common denominator is not this × this × this.1912

It is this × this because this factor is already contained in that.1917

Our lowest common denominator is x × x² + 6², and whatever is on top.1926

Be very careful, you are used to having the least common denominator, just multiply the denominators.1940

Here, because the factors are repeated, this does not need to be this × this × this.1945

It does not need to be that way, this is already contained in that.1951

It is just this and this.1953

Therefore, here what we need to do is we need to do a × x² + 6² + bx + c × x² + 6, only once.1955

And + dx + b × x/ x × x² + 6².1981

This denominator is the same as that denominator, which means the numerator is equivalent to the numerator.2003

Now we expand this numerator and that is what we are going to do next.2009

2 is going to equal, when I multiply all this out.2016

That is fine, it is just algebra.2025

ax⁴ + 12ax² + 36a +, it is going to end up being bx⁴ + cx³ + 6bx² + 6cx + dx² + e ⁺x.2030

When we combine terms, we have an x⁴ term.2070

This is going to be a, it takes care of that one, and a b.2077

It takes care of the x⁴ terms.2084

There is an x³ term.2087

The only x³ term is that one, c.2089

There is an x² term, 12a + 6b + d.2099

Make sure you get all the terms.2113

There is an x term, 6c + e.2116

There is a number term, 36a.2131

All of that is equal to 2.2138

We just set things equal to each other.2143

The equations that we get, in other words, a + b is going to be 0, c is going to be 0.2146

12a + 6b + d is going to be 0.2153

6c + e is going to be 0.2157

36a is going to equal 2, that is what we get.2160

We are going to get a + b = 0.2167

We are going to get c = 0, we are going to get 12a + 6b + d is equal to 0.2172

We are going to get 6c + e is equal to 0.2184

We are going to get 36a is equal to 2.2189

This gives us that a is equal to 1/18.2193

That takes care of that.2198

Let us go to the first one over here.2201

We use the equation a + b is equal to 0, which means that a = -b which means that b = -a which means of b = -1/18.2203

We already know that c is equal to 0.2222

We have taken care of that.2227

Now we have 6c + e is equal to 0.2229

c is equal to 0 so I get 0 + e is equal to 0, which means that e is also equal to 0.2237

Now I’m going to use this equation.2247

12 × 1/18 + 6b, 6 × -1/18 + d is equal to 0.2253

When I solve this, I get d is equal to -6/18.2267

I’m going to leave it as -6/18, instead of reducing that.2273

Therefore, our final partial fraction decomposition, our original function was 2/ x × x² + 6².2276

We said that was equal to a/x + bx + c/ x² + 6 + dx + e/ x² + 6².2289

Now we have a, b, c, d, and e.2304

Our decomposition is 1/18 / x + -1/18 x + c which is 0/ x² + 6 + -6/18 x + e which is 0/ x² + 6².2308

This is our final partial fraction decomposition.2336

That is just the decomposition, that is not the answer.2340

We still have to integrate this thing.2342

The integral of this is the integral of this because these are the same.2346

The integral of this is the integral of this + the integral of that + the integral of that.2351

Nice and simple, the rest is just using all the techniques that we gathered so far.2357

My final answer, our integral is going to equal 1/18 × the integral of 1/x dx - 1/182362

× the integral of x/ x² + 6 dx – 6/18 × the integral of x/ x² + 6² dx.2378

This is equal to 1/18 × natlog of the absolute value of x - 1/18 × ½, because of that u.2403

u = x² + 6, du = 2x dx, x dx = du/2 × natlog of x² + 6 – 6/18 × the integral of x/ x² + 6² dx.2417

How do we handle this?2444

We handle it this way, we do we a u substitution on this.2448

We let u = x² + 6, du = 2x dx, du/2 = x.2452

It is essentially the same thing that we did here, because of that extra x² part,2467

I thought it actually do the u substitution, x dx.2470

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 + 6.2478

Our final answer is = 1/18 × natlog of x – 1/18 × ½ × natlog of x² + 62509

– 6/18 × -1/ 2 × x² + 6 + c.2531

There we go, partial fraction decompositions are very tedious.2547

They are algebraically intense.2552

There are plenty of places where you can make a mistake but conceptually I do not think it is all together that difficult.2555

You just have to keep track of everything.2560

But this is calculus, you are more than accustomed to that by now because the problems just are,2562

by nature, sort of long and detailed.2569

Thank you so much for joining us here at www.educator.com.2574

We will see you next time, bye.2576