Welcome to a second video on integration
involving partial fraction decomposition.
I thought we should take a look at another example
where we have a quadratic factor.
So, let's go ahead and take a look at our integral.
To save a little bit of time, notice it's already
in factored form, but we have a linear factor
and a quadratic factor.
So, for the linear factor, we'll have a over x minus one.
And then for the quadratic factor, we'll have b x plus c
in the numerator.
Now we'll go ahead and multiply everything by the lcd,
which will be these two factors.
So on the left, both of these factors simplify out,
so we're left with five equals...
Here, the x minus one simplifies out.
We have a times x squared plus nine.
Here the x squared plus nine simplifies out,
so we have b x plus c, times x, minus one.
So, because of this quadratic factor,
we are going to have to multiply this out.
But before we do that, let's go ahead and use the zero
of this linear factor to determine the value of a.
So, if we let x equal one, we'll have five equals...
This will be one squared plus nine, that'd be 10 a.
And this would be zero, so we'd have a equals...
1/2.
Now, to find the values of b and c,
we will have to multiply this out
and then combine the like terms based upon the power of x.
So, we'll have five equals a x squared,
plus nine a,
plus b x squared,
minus b x,
plus c x,
and then minus c.
Let's identify the like terms based upon the powers of x.
So, here we have an x squared term and an x squared term.
Here we have an x term and an x term.
And these would be the constants.
So then five equals quantity a plus b,
times x squared,
plus negative b,
plus c x,
and then plus nine a, minus c.
Now, if we take a look at the left side of the equation,
we could write in here zero x squared,
plus zero x, plus five, if we needed to.
That just emphasizes that a plus b must equal zero.
Negative b plus c must equal zero.
And nine a minus c must equal five.
Remember, we already know that a is equal to 1/2.
So, if we take a look at this first equation,
if a is equal to 1/2, and a plus b is equal to zero,
we know that b is equal to negative 1/2.
And then looking at the second equation,
if negative b plus c equals zero,
well, negative b would be positive 1/2,
therefore c must equal negative 1/2.
So, way back up here in blue,
we can rewrite this integral now as the sum
of these two fractions based upon the values of a, b, and c.
Let's go ahead and do that on the next screen.
So, using these partial fractions,
now that we know the values of a, b, and c,
we can rewrite this integral
as a sum of these two fractions.
So, since a is equal to 1/2,
we would have 1/2 times the integral of one over x minus one
dx, now, since b and c are equal to negative 1/2,
I'm going to factor out a negative 1/2
and write minus 1/2 times the integral of x plus one
over x squared plus nine, dx.
Again, if I distribute the 1/2,
I'd have negative 1/2 x minus 1/2,
which match for the values of d and c.
Now, this is going to be a lot easier to integrate in this form
but this one is still going to be a little bit of a challenge.
Let's go ahead and rewrite this as two different integrals.
One will be x over x squared plus nine,
and the other will be one over x squared plus nine.
So, I'll have minus 1/2 times the integral of x
over x squared plus nine,
and then minus 1/2 times the integral of one
over x squared plus nine,
dx.
For this one, if we let u equal the denominator,
then du will equal dx.
So, this will be the antiderivative of one over u.
For the second one, if we let u equal the denominator
x squared plus nine,
differential u is going to equal two x d x.
So, if we divide both sides by two,
we can replace x d x with 1/2 du.
So we can use a basic integration formula for this as well.
And then for this last one, this should remind us
of the integral of one over a squared plus u squared.
So, using this basic integration formula here,
a is going to equal three,
and u will be equal to x.
So, I think we have everything.
Let's go ahead and try it.
Again, if this is equal to u, we'll have one over u, du,
so this is going to equal 1/2 times the natural log
of the absolute value of x minus one.
Now, here we're going to have minus 1/2,
and then when we perform the u substitution,
we're going to have another factor of 1/2,
because x d x is equal to 1/2 du.
So, we'll have 1/2, natural log,
absolute value of x squared plus nine.
And for this last integral, we'll have minus 1/2 times...
Again, we're using the integration formula here,
so we'll have one over a, which is 1/3, arc tangent
of u over a, while u is x and a is three.
So, we have x over three and then plus c.
Let's go ahead and clean this up one more time.
1/2 natural log, absolute value of x minus one,
minus 1/4, natural log of absolute value
of x squared plus nine.
This will be minus 1/6
arc tangent,
x over three, and then plus c.
So, even though partial fractions did allow us
to integrate this, it still was fairly involved.
But, it did allow us to find the antiderivative.
I hope you found this example helpful.
Have a good day.