Welcome to Part 1 of
Integration Involving Partial Fraction Decomposition.
Partial fraction decomposition is a process used
to rewrite rational functions as a sum or difference
of rational functions that are easier to integrate.
So in Algebra, we often combine rational expressions,
as we see here, by obtaining a common denominator,
but in Calculus, we often want to reverse the process
to help integrate.
So in Calculus, if we had this rational function,
it's going to be easier to integrate if we write it as a sum
of these two simpler rational functions.
And in most cases we use this integration formula here,
where the integral of one over u with respect to u
is equal to natural log absolute value u plus C.
So now we'll review the process
for performing partial fraction decomposition
and then take a look at two examples.
Before we perform partial fraction decomposition,
as an initial step, if the degree of the numerator
is greater than or equal to the degree of the denominator,
we must divide the denominator into the numerator
to obtain a polynomial plus a rational expression
where the degree of P sub one, this polynomial here,
is less than the degree of Q, our denominator.
So again, the if the degree of P is greater than or equal to
the degree of Q, we'll perform long division resulting
in a polynomial plus a rational expression where the degree
of P sub one will be less than the degree of Q.
So once the degree of the numerator
is less than the degree of the denominator
we can perform partial fraction decomposition.
And for Step One, we completely factor the denominator
into prime linear and quadratic factors.
Step Two, for every distinct linear factor,
we must include a fraction in this form here,
where the numerator is an unknown constant
and the denominator is a linear factor.
And for repeated linear factors in this form here,
notice how we have n repeated linear factors
we must include n terms in the form described here.
Where the numerator consists of unknown constants
and each fraction picks up one repeated linear factor
until we have all n repeated factors.
For quadratic factors in this form,
we must include fractions that look like this,
where now, the numerator consists of linear factors
and the denominator is our quadratic factor.
And once again, if we have repeated quadratic factors,
we must include fractions that look like this,
until we have all n repeated quadratic factors.
So once we have the partial fractions,
we multiply each fraction by the LCD
and the result is called the basic equation.
So the last step will be to solve the basic equation
by selecting convenient values of x
to determine the numerators of the partial fractions.
So let's take a look at our first example.
As a first step, notice the degree of the numerator is zero
and the degree of the denominator is two.
And since the degree of the numerator is less than
the degree of the denominator, we can go ahead
and perform partial fraction decomposition.
So for the first step, let's go ahead
and factor the denominator.
We'll have two binomial factors, we'll have x and x.
The factors of negative six that add to negative one
are negative three and positive two.
So now we'll take the rational function
and perform partial fraction decomposition.
So we'll have two divided by the quantity x minus three
times the quantity x plus two,
we have two distinct linear factors,
so our first fraction will contain an unknown constant A
divided by the first distinct linear factor of x minus three
this linear factor is not repeated,
so we go to the next linear factor of quantity x plus two
so we'll have plus B, another unknown constant,
divided by the quantity x plus two.
These are our two partial fractions
and now we're going to clear the fractions
from this equation by multiplying everything by the LCD,
which will give us our basic equation.
And since the least common denominator would be
the quantity x minus three times the quantity x plus two
we'll multiply every fraction by those two linear factors.
Notice how in this first fraction,
we have x minus three over x minus three
and x plus two over x plus two,
which just gives us two equals
this next product we have an x minus three
over x minus three, leaving us with A times
the quantity x plus two, plus here we have
an x plus two over an x plus two,
leaving us with B times the quantity x minus three.
We're now going to select convenient values of x
in order to solve for A and B.
For example, notice when x equals negative two,
the quantity x plus two would be zero,
so x equals negative two is
a convenient value of x to select.
So when x is negative two, we would have two equals,
this should be zero, and since x is negative two
we'd have B times negative five or negative five B.
Dividing both sides by negative five we can see
that B must equal negative 2/5.
So looking above for a moment,
for this fraction here, where we had B divided by
the quantity x plus two, B would be negative 2/5.
Now for the next value of x to select,
notice for the quantity x minus three,
if x equals three, this would be zero,
so now we'll select x equals three,
which would give us the equation two
equals, if x is three this would be A times five or five A,
and again this would be zero,
so dividing both sides by five,
we now know that A must be 2/5.
Which means our integral of two divided by
the quantity x minus three times the quantity x plus two
is equal to the integral of, where A is 2/5,
so we'd have 2/5 divided by the quantity x minus three
and since B is negative, it'd be minus 2/5 divided by
the quantity x plus two integrated with respect to x.
Let's go ahead and finish this on the next slide.
Let's go by this as two integrals
and we'll also factor out the 2/5.
So this is equal to 2/5 times the integral of one divided by
the quantity x minus three, and then minus 2/5 times
the integral of one divided by the quantity x plus two.
If we try to perform u substitution here,
notice u is equal to x minus three
and u is equal to dx, so these don't require u substitution,
this would just be 2/5 natural log absolute value
of the quantity x minus three,
minus 2/5 times natural log absolute value
of x plus two plus C, a constant of integration.
Let's go ahead and take a look at a second example.
The first thing we should notice about this example is
that the degree of the numerator is equal
to the degree of the denominator.
So we have to perform long division
before we perform partial fraction decomposition.
So let's go ahead and do this over on the side.
We have x squared plus three x minus four
divided by the quantity x squared minus two x minus eight.
To get started, we're only concerned about
this first term and this first term.
There's only one x squared in x squared,
so we'll have one up here,
we'll multiply and then subtract.
So one times this quadratic would just be
x squared minus two x minus eight.
The important thing to remember here is
that we are now going to subtract.
Instead of subtracting though,
we're going to add the opposite.
So we can change just the plus,
if we change this to a negative,
this to plus and this to a plus.
And now if we can just go ahead and add
this would be zero, this would be five x,
and this would be plus four,
which means the original integral is equal to
the integral of one plus the remainder of
five x plus four divided by the divisor of
x squared minus two x minus eight.
And now we'll perform partial fraction decomposition
on this fraction here.
Notice how the degree of the numerator is one
and the degree of the denominator is two.
So we can't forget about this one,
but now we're going to work
with just this rational expression.
So let's go ahead and factor the denominator,
so we have x and x, the factors of negative eight
that add to negative two are negative four and positive two,
again we have two distinct linear factors,
so we'll have A divided by the quantity x minus four
plus B times the quantity x plus two.
Now we're going to clear the fraction
by multiplying by the LCD,
which would be these two linear factors.
So now we'll multiply everything by
x minus four and x plus two.
Then we'll simplify.
So we have the quantity five x plus four equals
A times the quantity x plus two plus B
times the quantity x minus four.
This is our basic equation, so we'll take this over
to the next slide and solve for A and B.
And now we'll select convenient values of x.
So notice that x plus 2 would be zero
when x equals negative two.
And x minus four would be zero when x equals four.
So when x is negative two, this would be negative 10
plus four, that's negative six,
equals, this would be zero, this would be B times
negative six or negative six times B,
so B equals positive one.
And when x is four, this would be five times four plus four,
twenty four, equals six A and this would be zero,
dividing both sides by six, notice that A equals four.
So now that we know the values of A and B,
if we go back to the previous slide,
we'd have the integral of one plus these two fractions here,
with the values of A and B, that we just found.
So we'd have the integral of one plus A divided by
x minus four plus B divided by x plus two,
where now we know that A is four and B is one.
So you have the integral of one plus, A is four,
so four divided by the quantity x minus four, plus B is one.
We could separate this into three different integrals,
but let's go ahead and integrate in this form.
The integral of one would be x,
plus four times the integral of one
over the quantity x minus four,
would be natural log absolute value x minus four
plus the integral of one divided by the quantity x plus two
is natural log absolute value x plus two and then plus C.
So notice in these two examples,
all the denominators had linear factors.
In part two, we'll look at some examples
where we have quadratic factors as well as repeated factors.
Hope you found this first part helpful.