Hi, everyone! Welcome back to integralcalc.com. Today, we’re going to be doing another partial
fractions problem. And in this one, we’re going to be dealing with repeated quadratic
factors. We have a mix of factors. We have a linear factor which is x. we also have a
quadratic factor, x^2 + 1. And the fact that we have the quantity x^2 + 1^2 makes
this a repeated factor. Realize that if it’s x^2 + 1^2, it’s the same thing as (x^2 +
1)(x^2 + 1), which makes it repeated. This x^2 + 1 factor occurs once and then twice.
Since it occurs twice in our denominator, it’s a repeated factor. So dealing with
repeated factors makes this a little bit of a special problem. Whenever we’re doing
a partial fractions problem, we want to take our fraction exactly as it exists inside of
our integral and put it over here on the left-hand side of this equation that we’re going to
put together. So x^4 + 1 and then we take our denominator exactly as it is, x^2 + 1^2
and we put that on the left-hand side. Now, on the right-hand side, we’re going to start
developing the decomposition of this partial fractions problem.
So first we’re going to deal with the linear factor which is x. Since it’s a linear factor,
we just have to put a/x. Whenever you’ve got a linear factor, you just put the single
constant here, a/x. Then for the quadratic factor x^2 + 1, since it’s a repeated factor,
we have to treat it differently than a distinct factor or non-repeated factor. So remember
that if you’ve got a repeated factor for example, if you had a repeated factor x^4,
that’s repeated because it’s x(x)(x)(x). It’s actually repeated four times. In your
partial fractions decomposition, you have to include the factor x^4 but you also have
to include the factors x^3, x^2 and x or x^1. In this case, we have the repeated factor
x^2 + 1^2 so we have to include that as a factor but we also have to include (x^2 +
1)^1. So basically, if you have a repeated factor,
you have to include every factor of a lesser degree than the factor that you’ve been
given in the problem. So if we’ve got 4, we have to include 3, 2 and 1. If we’ve
got 2, we have to include 1 as well. So we need to include the factor x^2 + 1^2 and the
factor (x^2 + 1)^1. So since these are both quadratic factors, when you’ve go a linear
factor, you can just put the variable like this, a in the numerator, when you have a
quadratic factor, you have to put two constants in this variable in this form, Bx + C. So
you’ve got Bx + C and then we’d divide that by the first-order factor which is just
going to be (x^2 + 1)^1. Then we have the second quadratic factor. So we’ll put Dx
+ E and put that over the second quadratic factor which is x^2 + 1^2. So now we’ve
included both degrees of that repeated factor and we’ve got quadratic factors indicated
here in the numerator with these forms. So now that we’ve set up our decomposition,
we need to multiply both sides of this equation by the entire denominator of the left-hand
side. So we’re going to multiply every term on the left and the right-hand side by x(x^2
+ 1)^2. When we do that, obviously our entire denominator on the left-hand side will cancel
and we’ll just be left with x^4 + 1. On the right hand side, this x on the denominator
will cancel with this x, so we’ll be left with a times x^2 + 1^2. When we multiply this
Bx + C term by the denominator from the left, this x^2 + 1 term will cancel and we’ll
be left with (Bx + C)(x)(x^2 + 1)^1. Then when we multiply by dx + E, that entire x^2
+ 1^2 term will cancel so we’ll just get (Dx + E)(x). So now that we’ve multiplied
through everything, notice that the denominators on both sides of these equations have completely
disappeared. What we need to do now is use algebra to multiply out all of these things
together. And I’m going to go ahead and skip that part in this problem because the
algebra is just pretty tedious. But basically, what you need to do is multiply everything
out and the step you’ll get to when you multiply everything together is Ax^4 + 2Ax^2
+ A + Bx^4 + Bx^2 + Cx^3 + Cx + Dx^2 + Ex. So when you multiply everything out, this
is what you get. What you want to do now is go ahead and collect
like terms. So we’re going to collect all of our x^4 terms together, all of our x^3
terms together, x^2 terms together, etc. And I’ll just go ahead and cross these off as
I go so that we make sure we don’t miss any. So we have this a(x^4) and we have b(x^4).
So what we can do is actually if we collected those together, we would have Ax^4 + Bx^4.
If we factor out x^4, we’ll get (A + B)(x^4). And now these two are done. Now, if we collect
x^3 terms, there’s only one of them. So we’ll get plus cx^3. Now we want to look
at our x^2 terms. We’ve got 2Ax^2 and we have a bx^2 and a dx^2. Again, if we collect
those together and then factor out the x^2, we’ll get (2A + B + D)(x^2). Now if we collect
our x terms together, we’ll get Cx and Ex, again factoring out the x, we’ll get (C
+ E)(x). And then we just need to get our constants together and the only thing that’s
left is a. So we just go ahead and say, plus a. So that’s how we factor things out. And
the reason that we do that is because when we use this method for partial fractions,
in order for the left-hand side and the right-hand side of this equation to be equal to one another,
we know for example because the coefficient on the x^4 term on the left-hand side is 1.
We know that 1 has to be equal to the coefficient on the x^4 term on the right-hand side so
1 has to be equal to A + B. We also know that the constant over here on the left-hand side
1 has to be equal to a because a is the constant on the right-hand side. And now, we also know
that this coefficient c, this coefficient 2A + B + C, and this coefficient C + E all
have to be equal to zero since there are no x^3, x^2 or x terms on the left-hand side
of our equation. So given that, we can go ahead and write a series of simultaneous equations
so that we can solve for the variables Aa through E.
So again, we said that A + B was going to be equal to 1. We know that C = 0. we know
that 2A + B + D = 0. We know that C + E = 0 and we know that A = 1. So we want to go ahead
and solve for the variables A through E. And I’ll go ahead and write a legend up here
so that we can have a place to keep them. We already know that A = 1. We already know
that C = 0. If we go ahead and plug A = 1 into this equation, we’ll get 1 + B = 1.
Subtracting B from both sides gives us B = 0. So we know that B = 0. We can find E by plugging
C into this equation. So we know c = 0 so we get 0 + E = zero and we can see right away
that we’ll get E = zero. Then if we want to find D, we can go ahead and plug A and
B into this equation. A = 1 so we get 2 here. B = 0 so 2 + 0 + D = zero. So we’ll get
2 + D = zero which means that we’ll get D = -2. So those are our constants.
So now, we can take those and we can plug them back into our integral. So
we have the integral from 1 to 2. Instead of this nasty fraction here that we were given
in our original problem, we’re going to use everything over here on the right-hand
side, the decomposition along with the coefficients we just found and put that back into our integral
instead. So we know A is 1 so we get 1/x plus, B and C are both zero so we’ll get 0x +
0/(x^2 + 1). Then D and E are -2 and 0 respectively so we’ll get (-2x + 0)/(x^2 + 1)^2 dx. And
now this is actually something we can integrate. This entire second term here, obviously we’re
going to get zero in the denominator so this is going to go away. And what we’re left
with is the integral from 1 to 2 of 1/x, and I’m going to go ahead and split this apart
into two separate integrals so that we can easily deal with them one at a time. Then
we’re going to have a (-2x)/(x^2 + 1)^2 so we can go ahead and say minus 2 times the
integral from 1 to 2 of x over (x^2 + 1)^2 dx. So I just brought this negative two here
out in front of the integral and simplified it that way.
So now that we’ve got this, we can go ahead and integrate the first term easily. We know
that the integral of 1/x = ln |x|. So we’re going to evaluate that on the range 1 to 2.
And then for the second integral, we know that we can use u-substitution. We can see
it here in this x^2 + 1 term. So if we set u = x^2 + 1, and then if we take the derivative
to get du, we’ll get 2x dx. Solving for dx gives us dx = du/2x. so if we plugged that
substitution back into our integral, we’ll end up with x/u^2 and then we know that dx
= du/2x. So you can see that the x’s here are going to cancel. Let’s go ahead and
just continue doing this one step at a time. We’ll go ahead and evaluate the natural
log on the range 2 to 1. So when we do that, we’ll plug in 2 first and then subtract
and then plug in the lower limit of integration 1. So that’s taken care of. We’ve evaluated
that. Now we’ve got minus 2. And we can go ahead
and evaluate this 2 right here and bring it out in front of our integral. And since we
have minus 2 and we’re bringing this out in front; it’s in the denominator, we actually
end up with a minus 1. So those 2s cancel and we’re just with the integral from 1
to 2 of 1/u^2 du. And since we have 1/u^2, we can turn that into u^-2. Remember that
if you have 1/x^a, that’s the same thing as saying x^-a. You can move it into the numerator
by changing the sign on the exponent. So we’re going to go ahead and say that this is the
integral of u^-2 du. And that will make it easier for us to integrate. So now that we’ve
gotten this far, we’ll have the ln |2|. We know that ln |1| = 0. So we can go ahead
and get rid of this term. When we integrate u^-2, we’ll add 1 to the exponent so -2
+ 1 is a negative 1 but then we have to divide by our new exponent so our new exponent is
negative 1, so we need to divide by negative 1 like this and then we’re going to be evaluating
this on the range 1 to 2. So now, natural log of the absolute value
of 2, we’ve got a negative and a negative so this is going to be A + u^-1 and instead
of u^-1, we’ll move that back to the denominator so we get 1/u. we need to evaluate this on
the range 1 to 2. So now we know that u = x^2 + 1, we can go ahead and plug that back in
for u. So plus 1/(x^2 + 1). And again, we’ll need to evaluate from 1 to 2. So plugging
in our upper limit of integration 2, we’ll get 1/2^2 is 4 + 1 is 5, then we’ll subtract
and plug in the lower limit of integration so 1/1^2 is 1 + 1 is 2. So we’ve got that.
And now, we just need to find a common denominator here so we can use a common denominator of
10 which means we’re going to get 2/10. This is going to be 5/10. So we end up with
2/10 – 5/10 which is a negative 3/10. So we’ll get A + -3/10 which of course we can
just simplify to ln |2 - 3/10|. And that’s it. That’s our final answer.
So that’s how we use partial fractions with repeated quadratic factors to solve and get
our final answer. So that’s it. I hope this video helped you guys and I will see you in
the next one. Bye!