Welcome to calculus. 
I'm professor Ghrist. We're about to begin lecture 24 on 
partial fractions. >> The previous three lessons have 
solved integrals by means of running differentiation rules in reverse. 
We're going to change our perspective now and give an introduction to methods based 
on algebraic simplification. The broadest and more generally useful 
such method, is called the method of partial fractions. 
 >> In this lesson we'll look at certain algebraic methods for simplifying 
integrands. Indeed certain integrals respond well to 
a pre-processing step. This is especially true for rational 
functions ratios of polynomials for example, if you are asked to integrate 3x 
squared minus 5 over x minus 2. Then, naturally, you would perform the 
polynomial division and rewrite this as 3x plus 6 with a remainder of 7 over x 
minus 2. And while the former integral seems 
intimidating, the latter integral is simple 3x gives 3 halves x squared. 
6 integrates to 6x and 7 over x minus 2 yields 7 log x minus 2 and then add the 
constant. Now, this is a trivial example but what 
happens when you have a higher degree denominator in your rational integrand. 
We're going to rely on a certain result from algebra, which states the following. 
A rational function P of x over Q of x, where the degree of the denominator is 
greater than the degree of the numerator and for which the denominator has 
distinct real roots, r sub i. And that is, we're looking at an 
integrand of the form of polynomial P over x minus r1 times x minus r2, times 
etcetera, all the way up to x minus r sub n. 
Where Q has n distinct real roots, then this can be factored or decomposed as a 
sum of constants A sub i over quantity x minus r sub i. 
This fact is something that we're going to take for granted, we're not going to 
prove it but we're going to use it extensively. 
This integral of P over Q seems very difficult in general. 
However, the right hand side can easily integrated since all of these a sub i's 
are constants, then we get simply logarithms. 
Now, what do we do to come up with this decomposition? 
How do we determine these constants A sub i. 
This is sometimes called the method of partial fractions. 
Let's look at an example and simplify 3x minus 1 over x squared minus 5x minus 6. 
We can factor the denominator as quantity x plus 1 times quantity x minus 6. 
By our algebraic fact, we know that this must be A1 over x plus 1 plus A2 over x 
minus 6. For simplicity, let's call these 
constants A and B instead. And now to determine A and B, we're going 
to multiply out the right hand side, put it over a common denominator. 
Where we'll obtain a times quantity x minus 6 plus B times quantity x plus 1 in 
the numerator. And now we're going to expand that 
multiplication out and collect terms so, that the first degree coefficient is A 
plus B and the constant coefficient is B times 1 minus 6 times A. 
Now, these must match up with the coefficient of the numerator 3x minus 1, 
if we have two polynomials that are the same, then all of the coefficients must 
match up. And so we see we're reduced to two 
equations, in two unknowns mainly A plus B equals 3 and B minus 6A equals negative 
1. Now, we can solve such a linear system of 
equations for A and B. Might take a bit more work than I'm 
willing to put on this slide. So, let me show you a more direct method 
for computing A and B. If as before, we write out 3x minus 1 
equals A times x minus 6, plus B times x plus 1 then, instead of expanding and 
collecting, we can do the following. We can say, this is a true statement and 
it must be true for all values of x. Therefore, it must be true if x equals 
negative 1, and happens when we substitute in on the left? 
We get negative 4, on the right we get A times a negative 7 plus B times 0. 
And we have eliminated that variable from consideration. 
Solving, we obtained A is four seventh. Likewise, if we substitute x equals 6 
into the above equation, we get on the left hand side, 17. 
On the right hand side, A times 0 plus B times 7 we therefore, have eliminated A 
and we can conclude that B is seventeen sevenths. 
And I'll let you check that these two satisfy the two equations in the lower 
left hand corner. Now, lets put this method to work. 
Compute the integral x squared plus 2x minus 1 over 2x cubed plus 3x squared 
minus two x. Factoring the denominator gives us x, 
quantity 2x minus 1, quantity x plus 2 all multiplied together. 
Now, we hope to split these up into A over x plus B over 2x minus 1 plus C over 
x plus 2. Let's color code these and then multiply 
everything out and see what we get. A times 2x minus 1 times x plus 2, plus B 
times x times x plus 2, plus C times x times 2x minus 1 must equal the numerator 
of the integrand x squared plus 2x minus 1. 
Let's use our direct method of substitution. 
First we'll substitute in x equals 0. When we do that, we eliminate all of the 
terms except the one involving A. We obtain negative 2A equals minus 1. 
Solving, we get a is a half. Likewise, if we substitute in x equals 
one half, we eliminate all the terms except for the B, which we solve to get B 
as one fifth. What's our last root? 
If we substitute in x equals negative 2, then we obtain a value for C of negative 
one tenth. Substituting these coefficients back in 
to the integrand gives us for our integral, the integral of one half over x 
plus, one fifth over 2x minus 1, minus one tenth over x plus 2. 
These are all going to be in logarithms. We're going to get one half log of x, 
plus one tenth log of 2x minus 1 being careful with that one half coefficient, 
minus one tenth log of x plus 2. Let's put this to use in a differential 
equation. Consider the deflection x, as a function 
of time t associated to a thin beam that we've applied force to. 
Question is how does this deflection evolve? 
Well, if the load is proportional to a positive constant lambda squared, then 
one simple model for this beam would be dx dt equal lambda x squared minus x 
cubed. We can factor that as x times lambda 
minus x times lambda plus x. And now applying the separation method, 
we get on the left dx over x times lambda minus x times lambda plus x. 
On the right, dt. Integrating both sides prompts the use of 
a partial fractions method. I'll leave it to you, to show that the 
coefficients in front of the terms are respectively, 1 over lambda squared, 1 
over 2 lambda squared and negative 1 over 2 lambda squared. 
So, that when we perform the integral, we get a collection of natural logs with 
coefficients in front namely, log of x minus one half log of lambda minus x, 
minus one half log of lambda plus x equals lambda squared t plus a constant. 
Where I've moved the lambda squared's to the right hand side for simplicity. 
Now, what we would do next is try to solve for x as a function of t. 
This looks like it's going to involve a lot of algebra. 
And I'm not really in the mood so, let's use linearization and solve for the 
equilibrium to see if we can understand what's happening. 
If we plot x dot versus x, we see that there are three roots. 
One root at zero, one root at negative lambda and one root at positive lambda. 
By looking at the sign of x dot or by linearizing the right hand side of the 
equation, we can see that zero is going to be an unstable equilibrium. 
That makes physical sense, if you're compressing the beam if it were perfectly 
symmetric, then it would not deflect. However, that situation is inherently 
unstable and if you start off with just a little bit of deflection in either 
direction, you will quickly move to one of the stable equilibria at lambda or 
negative lambda. Depending on which side you buckle. 
Now, there are some complications involved with applying the method of 
partial fractions. We've assumed existence of real distinct 
roots. That is, a bit of a luxury, in general 
you don't have that. Sometimes you have repeated roots, 
something of the form of polynomial over x minus root r to the nth power. 
This decomposes but it decomposes into something a bit more complicated. 
There are n terms, each is a constant over x minus r to some power ranging from 
one up to N. Now all of these individual terms are 
Integrable, but some might take a little more work than others. 
What is potentially worse is complex roots, where you don't necessarily have 
real roots. Let's say, we look at a polynomial over a 
quadratic, ax squared plus bx plus c. Then if the discriminant b squared minus 
four ac is negative, we have complex roots. 
This does break up, but it breaks up into a linear polynomial in the numerator and 
then the quadratic in the denominator. And if you have repeated complex routes, 
well then it gets even more difficult still. 
All of these examples are doable, but very cumbersome. 
And I'm not going to test you on these in this course, in the end it's really just 
algebra. And in the end, it's really just roots, 
the roots control everything whether they're real, whether they're complex, 
whether they're distinct or repeated. As we saw in the example of the buckling 
beam, the roots have physical interpretations in differential equation 
models. Always pay attention to your roots, not 
just in calculus when trying to solve an integral, but in differential equations 
or even later still when you take linear algebra, you will see that the root 
influences everything. >> We've seen in this lesson how 
algebraic methods can be incredibly helpful. 
And incredibly complicated. We're not going to descend any further 
into that underworld of integration techniques. 
Rather, we're going to turn our gaze to a new character in our story. 
That, of the definite integral.