Welcome to the first of two videos
on integration involving partial fraction decomposition.
This video does assume you already have watched the videos
to learn how to perform partial fraction decomposition.
So let's go and take a look at our first example.
One of the main reasons why we might want to perform
partial fraction decomposition on this integrand
is to avoid having to use trigonometric substitution.
While trig substitution is a great method to integrate
when appropriate, it can get somewhat involved.
So the idea here is if we can rewrite this as a sum
or difference of a simpler rational function,
it would be much easier to integrate.
So let's start with our integrand and factor the denominator
it'd be x minus three times x plus two.
We have two linear factors, so we're going to have A
over x minus three plus B over x plus two.
Now we're going to multiply every term by the LCD x minus three
times x plus two.
So on the left these two factors simplify out so we're left
with two equals, here the x minus three simplifies out,
we're left with A times x plus two, plus,
here the x plus two simplified out, we're left
with B times x minus three.
This is our basic equation true for all values of x.
So we'll let x equal the zeros of the binomial factor.
So if x equals negative two, we'd have two equals
this would be zero plus B times negative five.
So dividing both sides by negative five we'd have B equals
negative two fifths.
Now let's let x equal positive three.
So we'd have two equals, this would be five A this would be
A times five, and this would be zero.
So we'd have A equals positive two fifths.
So using partial fraction decomposition we can rewrite
this integral as integral of A over x minus three
and A is two fifths.
So two fifths and one over x minus three dx,
plus B over x plus two.
Well since B is negative two fifths, we'll write this
as minus two fifths times the interval of one over x
plus two dx.
And in both cases here if we let u equal the denominator,
du will equal dx, so both of these would be the integral
of one over u, which is equal to natural log u.
So we have two fifths natural log of u which is,
if I have x minus three minus two fifths,
times the natural log of the absolute value of x plus two,
plus C.
So again the whole idea is if we take the original rational
function and perform partial fraction decomposition,
these new integrals will be much easier to integrate
than the original.
And as we see here they work.
Let's go and take a look at another example.
Now in this one we could take the numerator and divide
by the denominator, but let's go ahead and try to perform
partial fraction decomposition and see what we come up with.
Let's go ahead and factor the denominator.
We'd have x minus four, times x plus two, again,
both linear factors.
So we'd have A over x minus four,
plus B over x plus two.
Now I'll multiply every fraction by the LCD.
So on the left, all of this simplifies out, we're left
with x squared plus three x minus four,
must equal here, the x minus four simplifies out,
we'll end with A times x plus two.
Here the x plus two simplifies out
we have plus B times x minus four.
Again this is true for all values of x, but we'll select
the x values that would be zeros of the binomial factors.
Let's let x equal negative two.
So here we'd have four,
plus negative six, that'd be negative two minus four,
that would give us negative six,
must equal, this should be zero,
and this should be B times negative six.
So B is equal to one.
Now let's let x equal four.
If x is four, we'd have 16 plus 12 that'd be 28 minus four,
be 24 must equal, this should be A times six,
this would be zero, so A is equal to four.
So we can rewrite this original integral as a sum
or difference of two integrals based upon the values
of A and B from these partial fractions.
So since A is equal to four, we'd have four times
integral of one over x minus four, dx.
And since B is positive one, we'd have plus the integral
of one over x plus two, dx.
Again if you let our denominators equal u,
du is going to equal dx, so these would both be the integral
of one over u.
So we'd have four times a natural log of the absolute
value of x minus four, plus natural log of the absolute
value of x plus two, plus C.
So you can see that performing partial fraction
decomposition can come in handy
when integrating rational functions.
Okay in part two we'll take a look at a rational function
where we have quadratic factors.
I hope you found this helpful.