First of all, to integrate this, we have to make sure we do partial fractions
and then we integrate these two little fractions.
Do not just put on ln(𝑥²+𝑥) because this is a quadratic, right?
Anyway, factor this; we can take out 𝑥 and then we have 𝑥+1.
Now, let's go ahead and look at the integral from 1 to ∞.
For the first fraction, we must have some number over 𝑥,
plus the other fraction, some number over 𝑥+1, and then the 𝑑𝑥.
As you can see, these two are just linear factors, right?
So the top right here, there would just be numbers.
To figure this number out,
all we have to do is go back to the original and then cover up the same denominator.
And you have to ask yourself:
"How can you make this factor zero?"
Well 𝑥 has to be zero, right?
In that case, what you have to do is you put 0 into the other 𝑥.
And that's the cover-up method, so yeah.
1/(0+1) is, of course, just 1, and that's very nice.
Let's put 1 right over here.
Now, we look at this right here.
The denominator is 𝑥+1.
We go back to the original and cover this out, right?
And then we have to ask ourselves:
"How can we make 𝑥+1 equal to zero?"
In that case, 𝑥 has to be -1, and we put -1 into this 𝑥,
So we end up with 1/(-1), and, of course, that will be -1.
So that's how you do the cover-up method for the partial fraction.
And now this is so nice because we can just integrate, integrate, right?
Now, integral of 1/𝑥-- we get ln(|𝑥|),
and then this is the minus, so let's put that down.
And then, integrating this, (-1)/(𝑥+1),
just ln(|𝑥+1|), and we have the negative right here already, right?
And, yeah, some people would say you should put parentheses there;
I will put it down for you guys.
Anyway, that's the integral part.
Now, don't forget we still have to put this in the parentheses
and we have to go from 1 to ∞.
Here is the deal, though.
If we put ∞ to here, we will get ln(∞), which is ∞;
If you put ∞ to here, we get ln(∞+1).
The result, then, will also be infinity.
∞ - ∞, we have to do more work.
This is how we can handle this.
Because the inside here is just two natural log's,
We can combine them, right?
So, this right here inside is just natural log absolute value,
and the truth is, the absolute value doesn't really matter
because everything here is positive, but let me just put it down anyway.
Stay positive all the time, right?
Anyway, 𝑥 on the top,
and then since you are minus-ing, so it becomes division like this,
and then the 𝑥+1 inside.
So that's the combination of this
and then you just go ahead and plug in the numbers.
Keep in mind whenever we put ∞ trough here and here,
the idea is that we are taking a limit.
So technically we shall supply somewhere that
the limit as 𝑥 going to ∞ of ln(|𝑥/(𝑥+1)|).
Let me just put on this right here as a note for you guys.
To do this we can just do this inside out,
especially the inside is just a rational function,
and the easier way to do-- the easiest way to do this is pay attention to the top.
Let's try, honestly, if you're on the top. It's just 𝑥, and on the bottom, it's just 𝑥,
so yeah, 𝑥/𝑥 is just 1 so all in all, we get ln(1) and we get 0 for that.
So the idea is, when we put in ∞ here, we are taking a limit,
so we actually get the first part being 0 when we have infinity to here.
And then we have to put 1 into the 𝑥.
So we minus ln(|1/(1+1)|), like this.
Now, this right here, it's, of course 0 minus this,
which is negative natural log of-- this is 1 over 2, so that's ½, like this, absolute value.
Of course, absolute value doesn't matter anymore.
But the last thing is, this is just like -1.
You can put this to become a power here.
So you're looking at this as natural log-- let me use the parentheses now.
(½)⁻¹, like this.
And you can just work this out, because this is like saying
natural log-- (1)⁻¹ is 1, and then you have 1/((2)⁻¹),
And this is just a regular 2, right? So all in all, the answer is ln(2).
So, that's it, right? So that's it, that's it, that's it.
Now if you see, the improper integral here, it's a finite value.
That means this right here converges, because we just did a computation.
We can be sure that this right here also converges by the integral test.
Of course, you should also have to tell me that
this function here is positive, decreasing, and continuous on this interval.
But you'll just draw the picture on your own.
Anyway, we do know this right here also converges by the integral test.
And now I'm gonna show you guys what this is converging to, alright?
So here is the deal:
We are still going to do partial fractions;
I will factor out the 𝑛 this time, like this.
And of course, this is Σ as 𝑛 goes from 1 to ∞---
Let's open up the parentheses,
and for the first fraction,
we have some number over 𝑛 and then adding with some number over 𝑛+1, like this.
What's this number? Yes, it's 1. What's this number? Yes, it's -1. Why?
Because we did it over here already, right?
So it's the same kind of thing that we do.
But this time, though, we don't integrate them.
This right here is telling you that we are going to have an infinite sum.
Here is the deal:
We are going to put in 1 for the end first, and that's going to be the first parentheses.
It has 2 fractions.
The first 1/1, which is 1, and then we minus, all right? This is a minus.
And we put one right here, which is 1 over, 1+1, which is 2.
Congratulations, this is the first parenthesis.
And then the next one,
when 𝑛 is 2, put it here, put it here. We get ½, and then minus ⅓.
And then we pretty much just continue.
Next one, plus ⅓ when 𝑛 is 3, right? And then minus ¼.
And then so on, so on, so on.
And as you can see,
this right here is going to be a telescoping series
because in the middle, you see, this right here is -½.
This is a +½. Of course, they cancel.
Likewise, this right here is -⅓ can be canceled out with the next one very nicely.
So just keep canceling. Very nice.
As you can see, in the front, we have one remaining term
and the reason is because from 𝑛 to 𝑛+1,
The difference is 1, so you have one term in the front left,
and you also have to remember you have to write down the 𝑛th term for it.
And because we only have the difference being 1,
we just need to put on the 𝑛th term, and the 𝑛th term is just this right here, namely, 1/𝑛 - 1/(𝑛+1), like this.
And technically I should also put on plus, dot, dot, dot,
and it's because this right here keeps on going forever.
Now. If you recall from the second parentheses,
This guy got cancelled out with the number in front, right?
This guy got cancelled out with this number right in the front.
Now. If you look at this parentheses,
let me tell you this number is going to be cancelled out with the number right in front.
And so on, so on, so on.
But of course we are going to leave it right here.
You have to write on 𝑛th term because this right here is technically the 𝑛th partial sum formula.
And let's just write that down.
We will have 1, and this is minus 1/(𝑛+1).
The 𝑛th partial sum formula,
and then, what we do is we are going to take the limit,
because this right here, it keeps on going forever.
Take the limit as 𝑛 goes to infinity.
And be sure to do this step,
because sometimes this might contribute some negative number that you have to pay attention to.
But in this case, we have 1.
And, of course, -1/∞, this right here will approach 0,
so this is just minus 0, and all in all, the answer to this right here is equal to 1.
So, as you can see,
This improper integral converges to ln(2), which is about zero point...
Let me just write down for you guys, which is about 0.693...
And this infinite series converges to exactly one. Very nice.
And finally, just keep this in mind, either both of them converge or both of them diverge.
However, if both of them converge, they may not converge to the same value.
Most likely they will converge to a different value, like, this is an example for that.
So, keep this in mind.
And again, we can only calculate a certain sum, right? Just like this right here.
This is a telescoping series and this is how you're supposed to do it.
And you guys can check on my other videos