Hi, everyone!
Welcome back to integralcalc.com.
Today we’re going to be doing a partial fractions problem.
And in this particular problem, we’ve got the function x^2/(x + 2)^3.
And we’re going to be taking the integral of that and in order to do so, we’re going
to sue partial fractions.
The interesting thing about this problem is that we have linear repeated factors.
So remember that when you’re dealing with a partial fractions problem, the first thing
you need to do to decide is whether or not you’re dealing with linear factors or quadratic
factors and in either case, whether or not you’re dealing with distinct factors or
repeated factors.
In this case, in our denominator we don’t have any x^2, x^3, x^4 terms or any thing
like that.
We just have an x^1 power right here.
So that means that we’re dealing with a linear factor.
And because it’s cubed, because we have (x + 2)(x + 2)(x + 2), that linear factor
is repeated so we’ve got a linear repeated factor.
We’re going to treat that in a very specific way as opposed to distinct linear factors
or distinct quadratic or repeated quadratic factors.
So just remember that for partial fractions, what we’re going to do is we’re going
to take our function and we’re going to put it over here on the left-hand side of
the new equation that we’re going to create exactly as it is.
On the right-hand side, we’re going to go ahead and perform the partial fractions decomposition.
And the way that we’re going to do that with a linear repeated factor is we’re going
to make sure that we include every factor including it’s lesser degree.
SO we’re going to include (x + 2)^3 but we’re also going to include all lesser factors.
We’re also going to include (x + 2)^2 and (x + 2)^1 which will just be x + 2.
We don’t need the parenthesis.
So when we have linear repeated factors, we have to include the factor itself (x + 2)^3
and each of the lesser degree factors so in this case because it’s cubed, we have to
include (x + 2)^2 and (x + 2)^1.
If we had (x + 2)^4, we’d have to include that as well as all of these lesser factors.
And this is again something we do when we have a repeated factor.
So this is our partial fractions decomposition.
Now, we need to go ahead and multiply both sides of our equation by the denominator of
the left-hand side.
So essentially, we’re going to multiply both sides by this so we’re going to take
the entire equation and we’re going to multiply by (x + 2)^3.
So of course on the left-hand side, that’s going to cancel the denominator and we’ll
be left with x^2.
When we multiply it by this A/(x + 2)^3, it will also cancel the denominator and we’ll
just be left with A. When we multiply it by the term where the numerator is B, the denominator
will cancel and we’ll just be left with (x + 2)^1 in the numerator.
And when we multiply it by the last term, C/(x + 2), we’ll be left with (x + 2)^2
in the numerator.
So now that we’ve gotten rid of all our fractions and we need to go ahead and multiply
out the right hand side of the function.
So A + Bx + 2B and then we’ll go ahead and multiply this (x + 2)^2 so we’ll get x^2
+ 4x + 4 when we FOIL that out.
And we’re going to distribute that C. So we’ll end up with A + Bx + 2B + Cx^2 + 4Cx
+ 4C.
Now that we’ve multiplied everything out, we need to go ahead and group like terms together
so we’re going to group all of our x^2 terms together, all of our x terms together and
then all of our constant terms together.
And by constants we’re talking about terms that don’t involve the x variable.
The only term involving x^2 is the Cx^2 so we’ll go ahead and put the C in parenthesis
to illustrate a point that we’re going to make next.
So we’ve got Cx^2 and I’ll go ahead and cancel these out as we go.
Then for x^1, we’ve got Bx and 4Cx.
So we’ll say (B + 4C)x.
So I got Bx and 4Cx and I divided the x out of that term so I have (B + 4C)x.
Then we’re left with just constants so we’ve got plus A + 2B + 4C and these are all constants.
Now we can go ahead and say we’ve accounted for the 4C, the 2B and the A.
The reason we’re grouping terms like this is because what we’re going to do is we’re
going to equate coefficients on the left and right-hand sides of the equation.
So notice here that on the left-hand side, the coefficient on the x^2 is 1.
It’s 1x^2.
What we can do is we can go ahead and identify that the coefficient on the x^2 term is 1
on the left-hand side.
On the right-hand side, it’s C. which means that C must be equal to 1.
We also know on the left-hand side is x^1, 1x^2 which we already said but then it’s
+ 0x + 0 if we were going to go ahead and complete that polynomial.
So what we know is that in this case, B + 4C = 0 and we know that A + 2B + 4C = 0.
So we can go ahead and write those equations out so B + 4C = 0 and A + 2B + 4C = 0.
Now, we can use the method of simultaneous equations that we learned from algebra to
solve for A, B, and C.
We already know that C = 1 so let’s go ahead and plug that into the second equation because
as you can see, we’ll be able to solve for B. So we’ll get B + 4 = 0.
When we subtract 4 from both sides, we get -4.
So now we’ve solved for B, we can plug B and C into our third equation to find A. So
we have A + 2(-4) + 4(1) = 0.
And when we solve this, we will get A = 4.
So now we’ve solved for all four variables and we can go ahead and plug these back in
into our partial fractions decomposition up here and we’ll integrate that.
Instead of taking the integral of the original function now.
We’ll be taking the integral of 4, which we’re plugging in for A so 4/(x + 2)^3 then
we’ll plug in for B which is -4 so we’ll get – 4/(x + 2)^2 and then for C, we’ve
got 1 so we’ll have + 1/(x + 2) dx.
With partial fractions decomposition, now that we have three discrete fractions, we
can go ahead and break those apart into their separate integrals which will make this a
little bit easier for us to solve.
I’m also going to go ahead and bring out the 4 out in front of the integral as a coefficient
so what we’ll end up with is 4 times the integral of 1/(x + 2)^3.
Remember this rule from algebra that when you have 1/x^n or really any constant over
x^n, that’s the same thing as x^-n.
What this tells me is that when I have (x + 2)^3 in the denominator here, I can move
it to the numerator as long as I call it (x + 2)-3.
So I’ve got 4 times (x + 2)^-3 dx and then I’m going to move on to my second inte4gral.
So I’m going to say minus 4 times the integral of (x + 2)^-2 dx + the integral of, and we’ll
go ahead and leave this as it is because that’s easy to find the integral when we leave that
in the denominator.
So now we’re going to go ahead and integrate.
So here, remember that we’re adding 1 to the exponent so plus 1 gives us a -2 and then
we divide by our new exponent so I’m going to do 1/-2 (x + 2)^-2 and according to chain
rule, we’re going to have to divide by the derivative of the inside.
The derivative of (x + 2) is just 1 so I’m essentially just dividing by 1.
That’s all I need to do.
That’s going to be the integral of my first integral here.
Then I’ll have minus 4 times, again adding 1 is going to be -1 here in the exponent so
I’ve got 1/-1 (x + 2)^-1 and then finally, 1/(x + 2) is just that natural log of (x +
2) and of course I have to add c to account for my constant of integration.
Now, I’m just simplifying. 4/-2 gives me a -2 and I have a times (x + 2)^-2.
Remember that I can move that back to the denominator and make the exponent positive.
So I’ve got -2/(x + 2)^2.
-4/-1 gives me a +4.
And again I’m going to move (x + 2)^-1 to the denominator and that -1 becomes positive
1.
And then I’m going to go ahead and leave this as it is + ln (x + 2) + c.
The only thing I could do here, I could leave this as my final answer but it’s probably
cleanest if I simplify these first two pieces here into a common fraction.
So what I’m going to do is multiply this second term here by (x + 2)/ (x + 2) which
will give me my common denominator.
So what I’ll get is -2/(x + 2)^2 + 4(x + 2) on the numerator there give me 4x + 8/(x
+ 2)^2.
Now I can combine these and I’ll just go ahead and do that here.
So 4x + 8 – 2 is going to be (4x + 6)/ (x + 2)^2 and I’ve got to bring down my natural
log of (x + 2) and of course c, my constant of integration.
And that’s my final answer so [(4x + 6)/(x + 2)^2] + ln(x + 2) + c is the integral of
the original integral up here x^2/(x + 2)^3.
So that’s it.
I hope this video helped you guys and I will see you in the next one.
Bye!