00:01
Let's evaluate the following integral.
00:03
We should go ahead and check to see if this denominator, this quadratic term, will factor before we do partial fraction to composition.
00:11
So here we look at the discriminant b squared minus 4 ac.
00:17
In our case, that's just negative 4 squared then minus 4 times 1 times 6.
00:26
Now this is a negative number.
00:30
That means that the quadratic is irreducible.
00:32
It will not factor.
00:33
So using what the book calls case 4, our partial fraction decomposition should be of this following form.
00:46
And then we have another linear factor, cx plus d.
00:51
And then this time we have the same factor, but we'll square it.
00:56
All right, again, that's case 4.
01:00
Then let's go ahead and multiply both sides of this equation by the denominator on the left.
01:07
When we do that, and on the right, we have ax plus b, and then we have the, times the quadratic and then just cx plus d we can go ahead and expand that right side as much as we can bx squared and then combining depending on the power of x for example we could pull out x cubed we just have a so let me rewrite this and then pull out of x squared b minus 4a pull out of x and we have 6a minus 4b and we have 6a minus 4b plus c and then the constant term left over is 6b plus d and there's parentheses around that so now we look at the coefficients on the left and on the right on the left notice that there's no x cubed here this must mean that a is zero so then we also have b minus 4a that must equal 1 so since a is 0 we get b is 1 so these are two of our values.
03:04
Now let's plug these in for a and b over here.
03:09
And if we look at the left hand side, this should be equal to negative 3.
03:14
So we have negative 3 is 6a minus 4b plus c.
03:21
Now go ahead and solve that for c.
03:23
We get c equals 1.
03:26
And finally, let's solve for d.
03:28
6b plus d.
03:31
That's the constant term on the right.
03:34
That must equal 7.
03:35
The constant term on the left and then go ahead and solve that for d to get d equals 1 so now we have our four values a, b, c, and d let's go ahead and plug these in to the constants up here and then we'll take the integral of the right hand side let's go to the next page so plugging in our values for a, b, c, and d so this is our integral so let's go ahead and maybe let's flip this into two parts let's call this a and b.
04:36
So let's look at a first.
04:38
The first thing we should do for either of these integrals is complete the square.
04:42
So let's look at that quadratic.
04:47
We can go ahead and complete the square here and we'll end up with x minus 2 squared plus 2.
04:55
So let's look at part a first.
05:00
We have dx up top, x minus 2 squared.
05:05
And then i could write this as square root of 2 squared.
05:10
That'll make my choice for the truth.
05:12
Briggs sub more obvious, x minus 2 is root 2 tan theta.
05:20
Therefore, dx, square root 2, secan square theta, d theta.
05:28
Now let's go ahead and plug these in.
05:32
Route 2, secan square theta, d theta.
05:36
So we're replacing dx using this.
05:42
On the bottom, we have x minus 2 squared, so that will be this thing over here squared.
05:49
So that's 2 tangent square theta and then root 2 squared is also 2 and instead of writing that 2 there let me just put a 1 here and then i'll factor out that 2 in the front now recall tan squared plus 1 is equal to ccan squared so you could cross those off we have root 2 over 2 and then the ccans cancelled we just have integral d theta that's just theta and we could go ahead and solve for theta by using the tricks up.
06:29
So this, we can rewrite this as tangent equals x minus 2 over square root, then solve for theta by taking arc tan on both sides.
06:50
So plugging that in for theta, and let me not worry about the constant c because i still have to deal with this other integral b.
06:58
I'll add in the c at the very end.
07:05
And then we have x minus 2, radical 2.
07:10
So that takes care of our first integral a...