- [Instructor] All right, let's see if
we can find the indefinite integral of
one over five x squared minus 30x plus 65 dx.
Pause this video and see if you can figure it out.
All right, so this is going to be an interesting one.
And it'll be a little bit hairy,
but we're gonna work through it together.
So, immediately you might try multiple
integration techniques and be hitting some walls.
And what we're going to do here is actually try
to complete the square in this denominator right over here.
And then by completing the square, we're gonna get it
in the form that it looks like the derivative of arctan.
And if that's a big hint to you, once again
pause the video and try to move forward.
All right, now let's do this together.
So I'm just gonna try to simplify this denominator
so that my coefficient on my x squared term is a one.
And so I could just factor a five out of the denominator.
If I did that, then this integral will become
1/5 times the integral of one over,
so I've factored a five out of the denominator
so it is x squared minus six x plus 13 dx.
And then as I mentioned I'm gonna
complete the square down here.
So let me rewrite it, so this is equal to 1/5
times the integral of one over,
and so x squared minus six x is clearly
not a perfect square the way it's written.
Let me write this plus 13 out here.
Now what could I add and then I'm gonna
to have to subtract if I don't want
to change the value of the denominator.
In order to make, In order to make
this part right over here a perfect square.
Well we've done this before,
you take half of your coefficient here,
which is negative three and you square that.
So you want to add a nine here.
But if you add a nine then you have
to subtract a nine as well.
And so this part is going to be
x minus three squared and then this part
right over here is going to be
equal to a positive four and we of course
don't want to forget our dx out here.
And so let me write it in this form.
So this is going to be equal to
1/5 times the integral of one over,
get myself some space, x minus three squared
plus four, which we could also write as plus two squared.
Actually let me do it that way, plus two squared dx.
Now many of y'all might already say
hey this looks a lot like arc tangent,
but I'm gonna to try to simplify it even more
so it becomes very clear that
it looks like arc tangent is going to be involved.
I'm actually gonna do some
u substitution in order to do it.
So the first thing I'm going to do is
let's factor a four out of the denominator here.
So if we do that, then this is going
to become 1/5 times 1/4 which is going
to be 1/20 times the integral of
one over x minus three squared
over two squared.
And this is going to be a plus one,
and of course we have our dx.
And then we could write this as,
and I'm trying to just do every step here.
A lot of these you might have
been able to do in your head.
One over, and I'll just write this as,
x minus three over two squared plus one.
Plus one and then dx.
And now the u substitution is pretty clear.
I'm just going to make the substitution
that u is equal to x minus three over two.
Or we could even say that's u is equal to
one half x minus three halves,
that's just x minus three over two.
And du is going to be equal to one half dx.
And so what I can do here is, actually
let me start to re-engineer this integral a little bit
so that we see a one half here.
So if I make this a one half
and then I multiply the outside by two.
So I divide by two multiply by two.
It's one way to think about it.
This becomes 1/10 and so doing my u substitution
I get 1/10, that's the 1/10 there times the integral of.
Well I have one half dx right over here
which is the same thing as du.
So I can put the du either in the numerator
or I could put it out here.
And then I have one over,
this is u squared, u squared plus one.
Now you might immediately recognize,
what's the derivative of arctan of u?
Well that would be one over u squared plus one.
So this is going to be equal to 1/10
times the arc tangent of u, and of course
we can't forget our big constant C
because we're taking an indefinite integral.
And now we just want to do the reverse substitution.
We know that u is equal to this business right over here.
So we deserve a little bit of a drum roll.
This is going to be equal to 1/10
times the arc tangent of u.
Well u is just x minus three over two,
which could also be written like this.
So arctan of x minus three over two and then plus C.
And we are done.