Welcome to an example of integration
using partial fraction decomposition.
Before applying partial fraction decomposition
though, we should verify that
basic u substitution won't work.
Notice in this case,
the numerator has degree three
and the denominator has degree two.
While it might be tempting to
let u equal the higher degree part
or the numerator
notice a differential u
would be nine x squared
plus eight x plus one
times dx
and this does not fit the form of our integral.
Next before we apply partial fraction decomposition,
notice how the degree of the numerator
is not less than the degree of the denominator.
And when performing partial fraction decomposition
if the degree of the numerator
is greater than or equal to
the degree of the denominator
we must perform long division first.
So let's start there.
We would have three x cubed,
plus four x squared,
plus x minus six
divided by
x squared minus one.
Now to begin
we want to know what times x squared
would be equal to three x cubed
which would be three x.
So put three x over the x term,
multiply this term and the divisor.
So three x times x squared
would be three x cubed
three x times negative one would be
negative three x
which we will put under the x term here
and performing long division
we now need to subtract this
but instead of subtracting
we are actually going to add the opposite instead
so we can change this to addition
as long as we change the sign
of the two terms inside the parenthesis.
So if I change this to plus,
this would be negative
and this would become plus.
So notice how now,
so now when we add
instead of subtract
let's go and bring down
the four x squared
the x term would be four x.
Let's also bring down the minus six.
Now for the next term of the quotient
we determine what times x squared
would give us four x squared
and that will be positive four.
So the next term is plus four
and now again we are going to multiply
this term in the quotient and the divisor
so four times x squared is four x squared.
Four times negative one would be minus four.
Again we will subtract
by adding the opposite instead.
So we will change the sign here to plus.
Change the sign here and change the sign here.
Notice when we add down,
we bring the 4x down
and this would be minus two.
It is okay to have a degree one remainder
because our divisor is degree two.
What we've discovered so far
is this integral
can be written as the integral of
three x plus four
plus a remainder which we will write as
the quantity four x minus two
divided by the quantity x squared minus one.
So now we can easily integrate
the first two terms
but we are still left with this rational expression.
So again we might try use substitution
by letting u equal x squared minus one
and then differential u would be
two x dx
and so while we could actually use this substitution,
we broke this fraction up into
two separate fractions.
Meaning we could write this as
four x divided by the quantity x squared minus one
minus two divided by the quantity
x squared minus one,
so this u substitution
would help us integrate
this part here
but not the second fraction.
So what we are going to do is
go ahead and leave it in this form here
and then perform partial fraction decomposition
on this rational expression.
Let's go ahead and do that
on the next slide.
No we are going to take this ration expression here
and where it is a sum or difference
of fractions
that will hopefully be easier to integrate.
So to begin let's rewrite this
and factor the denominator.
So we would have the quantity four x minus two
divided by this is a difference of squares.
So we have x plus one times
quantity x minus one.
Because our denominator contains
two distinct linear factors
we are going to write this
as a sum of two separate fractions
but will have a constant
divided by the first linear factor
plus a constant divided by the
second linear factor.
Meaning we will have
A divided by the quantity x plus one
plus B times the quantity x minus one.
So once we determine the values
of A and B,
we can rewrite this rational expression as this sum.
So for the next step,
we are going to clear the denominators from this equation
by multiplying both sides of the equation
by the least common denominator
which would be the quantity x plus one
times the quantity x minus one.
Notice on the left side,
x plus one over x plus one
and x minus one over x minus one
simplify out
leaving us with just four x minus two
equals on the right side
notice in this first product here
one factor of x plus one
simplifies out
leaving us with A
times the quantity x minus one
and then for the second product here,
one factor of x minus one
simplifies out
leaving us with plus B
times the quantity x plus one.
This is called the basic equation
which we now want to solve
for A and B
by selecting convenient values of x.
For example notice if we let x equal positive one
this product here would be zero
so if x equals one,
we will substitute one for x
that would be four minus two or two equals
again if x is one
this should be zero
plus B times the quantity one plus one
or two B.
Notice here we can determine that
B equals one.
And have to determine the value of A
we will select x equals negative one,
which would make this factor zero.
Therefore this product zero.
So if x equals negative one
this should be negative four
minus two that's negative six
equals this would be
A times negative two or negative two A.
And again this would be zero.
Notice how here if were divide both sides
by negative two,
we can determine that A equals three.
So now if we substitute one for B,
and three for A,
we can rewrite this rational expression
as this sum,
so we can write this integral here
as the integral of
three x plus four
plus since A is equal to three,
we would have three divided by
the quantity x plus one
plus since B is one,
we'd have one divided by the quantity x minus one.
So again the whole point of doing this is,
going back to the visual integral.
We weren't able to integrate this
in its current form.
So we performed long division
and then to help us integrate this piece here
we performed partial fraction decomposition
which gave us a sum of these two rational expressions
and hopefully now in this form
we can now integrate.
Let's go ahead and do this
on the next slide.
The anti-derivative of three x
would be three times x to the second
divided by two
plus the anti-derivative of four
would just be four x.
To find this anti-derivative
we might be thinking
we need to perform u substitution
but notice if u is equal to
x plus one,
then differential u would just be equal to dx.
So this fits the form of the integral
of one over u du
which is equal to natural log
absolute value of u plus C.
Because we have a three
in the numerator
this would be plus three
times natural log
absolute value of x plus one
and for this term here,
we'd have plus natural log
absolute value of x minus one
and of course plus C.
Let's equate this one more time.
We would have three halves x squared
plus four x
plus three natural log
absolute value of x plus one
plus natural log
absolute value of x minus one
plus c.
This would be our anti-derivative.
I hope you found this helpful.