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[MUSIC]. 

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We already know how to integrate. 
And by integrate, I mean, 

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anti-differentiate some fractions. 
For instance, here is an 

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anti-differentiation problem. 
Let's anti-differentiate 4x plus 3 dx. 

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And I can do that with a u substitution. 
we have to choose what u should be equal 

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to. 
I'll set u equal 4x plus 3, so that du 

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is, what's the derivative of this, is 
just 4, and then the differential of u is 

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4dx. 
Now, I don't see a 4dx in my integrand 

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here, but I can manufacture a 4dx. 
If I multiply by 1 4th and 4. 

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That's like doing nothing, right? 
Those cancel, but now I've got a 4dx in 

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my integrand. 
So, the original anti-differentiation 

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problem is 1 quarter of anti-derivative 1 
over u du. 

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And I know an anti-derivative 1 over u 
du. 

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Alright? 
It's log of the absolute value of u plus 

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c. 
And, I've got to make sure to include 

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that constant on the outside of 1 
quarter. 

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But I don't want to write down my answer 
in terms of u, I'd like my answer to be 

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in terms of x. 
Right? 

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I'm asking for an anti-derivative 1 over 
4x plus 3, not somethings in terms of u. 

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So, I'll just replace u with what it's 
equal to. 

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So, altogether, I got 1 quarter log of 
the absolute value of 4x plus 3, plus c. 

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So, there's my anti-derivative of this 
fraction. 

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I also know how to anti-differentiate 
some reciprocals of a quadratic 

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polynomials. 
At least, here's a special case what's 

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the anti-derivative 1 over x squared plus 
one dx? 

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Well, if you remember back, I know a 
function whose derivative is this. 

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Alright. 
It's the inverse tangent function arctan. 

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So, at least I know how to 
anti-differentiate the reciporical of 

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this particular quadratic polynomial. 
Wonderfully, I can use that special case, 

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the anti-derivative of 1 over x squared 
plus 1, to anti-differentiate the 

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reciporicals of of other quadratic 
polynomials. 

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For example, can we find an 
anti-derivative of 1 over x squared plus 

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4x plus 7 dx? 
Can you think of a function whose 

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derivative is this? 
I can't come off with an anti-derivative 

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just off the top of my head. 
But, I've got a trick up my sleeve, the 

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trick is completing the square. 
What that means is that I'm going to 

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rewrite x squared plus 4x plus 7 and I 
want it to look more like the example I 

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know, right? 
The example that I know is this thing. 

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I know that the anti-derivative of 1 over 
something squared plus something. 

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So, I'm just trying to rewrite this 
quadratic polynomial, so it looks more 

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like that. 
So, I'm going to try to rewrite it, so it 

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looks like x plus, you know, something 
squared plus plus something. 

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The trouble is how to find the blue and 
the red somethings. 

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Let's do some algebra. 
So, I'm going to write this as x plus, 

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instead of just a squiggle, I'll call it 
a,squared plus some other constant b. 

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And if I expand that out, what do I get? 
Well, I get x squared plus a cross term 

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2ax, plus I got an a squared term. 
and then a b left over that I add on. 

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So, I'm trying to make this quadratic 
polynomial look like this. 

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And the only term that I've got with an x 
here is this 2a should be 4. 

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Alright? 
So, let me write that down. 

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So, 2a is equal to 4. 
And that tells me what a is. 

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a should be 2. 
I'll finish by finding b. 

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So, I've got x plus 2 squared plus some b 
is equal to x squared plus 4x plus 7. 

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x plus 2 squared is x squared plus 4x 
plus 2 squared, which is 4, plus b, is 

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suppose to be equal to this. 
And, that's tells me what b had better 

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be. 
Right? 

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I've got 4x and I've got an x squared, 
then I've got a 4 plus b is 7. 

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Alright, so 4 plus b is 7, so b is 3. 
Let's summarize. 

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So, what I've got here is that x squared 
plus 4x plus 7 can instead be written as 

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x plus, in this case, 2 squared plus, and 
then for b, I've got 3. 

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So, I've rewritten this quadratic 
polynomial as a linear term squared plus 

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some constant. 
Now, I'll use that fact in the 

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integration problem. 
Right, the original integral is 1 over x 

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squared plus 4x plus 7. 
And instead of doing that integral, I can 

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do the integral of 1 over this equivalent 
thing, x plus 2 squared plus 3 dx. 

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That still doesn't exactly look like the 
anti-derivative problem that resulted in 

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arctan. 
Remember, the anti-derivative here that 

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gave us arctan x has a plus one, but the 
thing we're looking at here has a plus 3. 

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So, I can fix that. 
I can get a constant here. 

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I can factor out something. 
I'll factor out a third and that will be 

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integral of 1 over 1 over 3 times x plus 
2 squared plus 1 dx, right? 

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This is the same thing as this, because 
this 3 times a third is just the 1 here 

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that I haven't written. 
And 3 times plus 1 gives me this plus 3. 

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I'll make a substitution. 
Now, I'll make this substitution. 

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you might think that I could use a 
substitution where u equals 1 3rd x plus 

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2 squared. 
But then, I need an x in the numerator, 

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right? 
So instead, I'm going to make it slowly 

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different substitution, I'm just going to 
make this looks like u squared. 

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So, I'll set u to be 1 over the square 
root of 3 times x plus 2. 

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So now, this denominator looks like 1 
over u squared plus 1. 

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And in that case, what's du? 
Well, that means du is 1 over the square 

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root of 3 dx. 
Now, what does the integral become? 

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You might be concerned that you don't see 
1 over the square root of 3 dx here, but 

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I can manufacture one of those. 
If I make this into the 1 over the square 

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root of 3, I can put a 1 over the square 
root of 3 here, without affecting 

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anything. 
And now, I've got a du. 

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So, this anti-differentiation problem 
becomes 1 over the square root of 3. 

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The anti derivative of 1 over this, here, 
is u squared plus 1. 

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And this 1 over the square root of 3 dx, 
that is now my du. 

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And that's arctan. 
So, this is one over the square root of 

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3. 
Anti-derivative of this is arctan u plus 

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c. 
Now, I just have to replace that u with 

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its value in terms of x. 
So, this is 1 over the square root of 3 

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arctan, and u here is this, so I'll write 
that in here for u, 1 over the square 

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root of 3 times x plus 2 plus C. 
And now, I'm in a position to state a 

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conclusion. 
The original question was asking for an 

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anti-derivative of 1 over x squared plus 
4x plus 7, and we found it. 

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It's 1 over the square root of 3 times 
arctan, 1 over the square root of 3 times 

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x plus two, plus some constant. 
So, we've anti-differentiated this 

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reciprocal of this quadratic polynomial. 
Well, fantastic, and the same trick works 

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in other situations as well. 
Really, anytime you've been handed a 

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quadratic polynomial but you wish you'd 
been given something squared plus a 

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number, you should try completing the 
square. 

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