[MUSIC]. 
We already know how to integrate. 
And by integrate, I mean, 
anti-differentiate some fractions. 
For instance, here is an 
anti-differentiation problem. 
Let's anti-differentiate 4x plus 3 dx. 
And I can do that with a u substitution. 
we have to choose what u should be equal 
to. 
I'll set u equal 4x plus 3, so that du 
is, what's the derivative of this, is 
just 4, and then the differential of u is 
4dx. 
Now, I don't see a 4dx in my integrand 
here, but I can manufacture a 4dx. 
If I multiply by 1 4th and 4. 
That's like doing nothing, right? 
Those cancel, but now I've got a 4dx in 
my integrand. 
So, the original anti-differentiation 
problem is 1 quarter of anti-derivative 1 
over u du. 
And I know an anti-derivative 1 over u 
du. 
Alright? 
It's log of the absolute value of u plus 
c. 
And, I've got to make sure to include 
that constant on the outside of 1 
quarter. 
But I don't want to write down my answer 
in terms of u, I'd like my answer to be 
in terms of x. 
Right? 
I'm asking for an anti-derivative 1 over 
4x plus 3, not somethings in terms of u. 
So, I'll just replace u with what it's 
equal to. 
So, altogether, I got 1 quarter log of 
the absolute value of 4x plus 3, plus c. 
So, there's my anti-derivative of this 
fraction. 
I also know how to anti-differentiate 
some reciprocals of a quadratic 
polynomials. 
At least, here's a special case what's 
the anti-derivative 1 over x squared plus 
one dx? 
Well, if you remember back, I know a 
function whose derivative is this. 
Alright. 
It's the inverse tangent function arctan. 
So, at least I know how to 
anti-differentiate the reciporical of 
this particular quadratic polynomial. 
Wonderfully, I can use that special case, 
the anti-derivative of 1 over x squared 
plus 1, to anti-differentiate the 
reciporicals of of other quadratic 
polynomials. 
For example, can we find an 
anti-derivative of 1 over x squared plus 
4x plus 7 dx? 
Can you think of a function whose 
derivative is this? 
I can't come off with an anti-derivative 
just off the top of my head. 
But, I've got a trick up my sleeve, the 
trick is completing the square. 
What that means is that I'm going to 
rewrite x squared plus 4x plus 7 and I 
want it to look more like the example I 
know, right? 
The example that I know is this thing. 
I know that the anti-derivative of 1 over 
something squared plus something. 
So, I'm just trying to rewrite this 
quadratic polynomial, so it looks more 
like that. 
So, I'm going to try to rewrite it, so it 
looks like x plus, you know, something 
squared plus plus something. 
The trouble is how to find the blue and 
the red somethings. 
Let's do some algebra. 
So, I'm going to write this as x plus, 
instead of just a squiggle, I'll call it 
a,squared plus some other constant b. 
And if I expand that out, what do I get? 
Well, I get x squared plus a cross term 
2ax, plus I got an a squared term. 
and then a b left over that I add on. 
So, I'm trying to make this quadratic 
polynomial look like this. 
And the only term that I've got with an x 
here is this 2a should be 4. 
Alright? 
So, let me write that down. 
So, 2a is equal to 4. 
And that tells me what a is. 
a should be 2. 
I'll finish by finding b. 
So, I've got x plus 2 squared plus some b 
is equal to x squared plus 4x plus 7. 
x plus 2 squared is x squared plus 4x 
plus 2 squared, which is 4, plus b, is 
suppose to be equal to this. 
And, that's tells me what b had better 
be. 
Right? 
I've got 4x and I've got an x squared, 
then I've got a 4 plus b is 7. 
Alright, so 4 plus b is 7, so b is 3. 
Let's summarize. 
So, what I've got here is that x squared 
plus 4x plus 7 can instead be written as 
x plus, in this case, 2 squared plus, and 
then for b, I've got 3. 
So, I've rewritten this quadratic 
polynomial as a linear term squared plus 
some constant. 
Now, I'll use that fact in the 
integration problem. 
Right, the original integral is 1 over x 
squared plus 4x plus 7. 
And instead of doing that integral, I can 
do the integral of 1 over this equivalent 
thing, x plus 2 squared plus 3 dx. 
That still doesn't exactly look like the 
anti-derivative problem that resulted in 
arctan. 
Remember, the anti-derivative here that 
gave us arctan x has a plus one, but the 
thing we're looking at here has a plus 3. 
So, I can fix that. 
I can get a constant here. 
I can factor out something. 
I'll factor out a third and that will be 
integral of 1 over 1 over 3 times x plus 
2 squared plus 1 dx, right? 
This is the same thing as this, because 
this 3 times a third is just the 1 here 
that I haven't written. 
And 3 times plus 1 gives me this plus 3. 
I'll make a substitution. 
Now, I'll make this substitution. 
you might think that I could use a 
substitution where u equals 1 3rd x plus 
2 squared. 
But then, I need an x in the numerator, 
right? 
So instead, I'm going to make it slowly 
different substitution, I'm just going to 
make this looks like u squared. 
So, I'll set u to be 1 over the square 
root of 3 times x plus 2. 
So now, this denominator looks like 1 
over u squared plus 1. 
And in that case, what's du? 
Well, that means du is 1 over the square 
root of 3 dx. 
Now, what does the integral become? 
You might be concerned that you don't see 
1 over the square root of 3 dx here, but 
I can manufacture one of those. 
If I make this into the 1 over the square 
root of 3, I can put a 1 over the square 
root of 3 here, without affecting 
anything. 
And now, I've got a du. 
So, this anti-differentiation problem 
becomes 1 over the square root of 3. 
The anti derivative of 1 over this, here, 
is u squared plus 1. 
And this 1 over the square root of 3 dx, 
that is now my du. 
And that's arctan. 
So, this is one over the square root of 
3. 
Anti-derivative of this is arctan u plus 
c. 
Now, I just have to replace that u with 
its value in terms of x. 
So, this is 1 over the square root of 3 
arctan, and u here is this, so I'll write 
that in here for u, 1 over the square 
root of 3 times x plus 2 plus C. 
And now, I'm in a position to state a 
conclusion. 
The original question was asking for an 
anti-derivative of 1 over x squared plus 
4x plus 7, and we found it. 
It's 1 over the square root of 3 times 
arctan, 1 over the square root of 3 times 
x plus two, plus some constant. 
So, we've anti-differentiated this 
reciprocal of this quadratic polynomial. 
Well, fantastic, and the same trick works 
in other situations as well. 
Really, anytime you've been handed a 
quadratic polynomial but you wish you'd 
been given something squared plus a 
number, you should try completing the 
square. 
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