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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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