00:01
In this problem, we want to evaluate the integral of 2y to the fourth over y -cubed minus y squared plus y -1 -di -y.
00:15
So in order to do this, we are going to have to divide the numerator into the denominator.
00:24
Sorry.
00:25
So we have to divide y cubed minus y squared plus y minus one into the two y to the fourth.
00:36
Because we want to break this fraction up into lower power fractions that we can just use definitions for like the integral of one over y and stuff like that.
00:48
So in order to do that, we need to use partial fractions to break it up.
00:51
But in order to use partial fractions, we need to use long division.
00:54
To get this into a more manageable form.
00:57
So i have y -cubed minus y -squared plus y -minus -y -squared plus y -fice -1.
01:00
I'm going to divide that into 2 -y -to -the -fourth plus 0 -y -squared plus 0 -y plus 0.
01:11
Okay, so first we're going to multiply by 2y.
01:14
So that's here.
01:16
So i'll have 2 -y to the 4th minus 2y squared plus 2y -minus 2 -2.
01:25
Something's wrong with my formatting here.
01:30
Yeah, this isn't right.
01:31
So it should be 2y cubed plus 2y squared minus 2y.
01:38
And when we subtract that, we're going to get 0, 2y cubed, minus 2y squared plus 2y.
01:49
And so now we're going to want to multiply this by 2, so plus 2.
01:57
So we have 2y cubed minus 2y squared plus 2y minus 2y minus 2.
02:05
And we'll subtract that.
02:06
We'll have the remainder as 2 because we had 0, 0, 0.
02:12
And we're left with the negative of negative 2.
02:16
Okay, so let's rewrite our integrand.
02:19
So we have 2y plus 2 plus 2 over yq.
02:26
Cubed minus y squared plus y minus one, d .y.
02:32
Okay, so we need to factor this denominator of this fraction in order to use partial fractions.
02:41
So we can group it.
02:42
So we have y cubed plus y minus y squared minus one.
02:48
I can rewrite this as y times y squared plus one minus y squared plus one...