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.