00:01
For this problem, we're going to go over example number two, which is to evaluate the integral 6x plus 7 divided by x plus 2 squared dx using partial fractions.
00:14
Now you can see that the first step is going to be splitting up the integrand using the partial fractions rule.
00:21
So one thing that's notable here is that you'll notice we have an x plus 2 squared in the original denominator.
00:27
Denominator because of that i'm going to need to split it into two separate terms one with just x plus two in the denominator but then i need another term with an x plus two squared in the denominator.
00:40
So this is going to be true any time i have a squared term in the denominator of a partial fraction decomposition.
00:50
So you can see here that my original integrand is 6x plus 7 divided by x plus 2 squared.
01:01
So as you can see in the example, we're going to split that up into a divided by x plus 2 and b divided by x plus 2 squared.
01:11
Now you can see the next step is going to be to solve for a and b.
01:16
In order to do this, we're going to start by multiplying both sides of the equation by the denominator of the left hand side, which is x plus 2 squared.
01:26
So we're multiplying everything in the equation by x plus 2 quantity squared.
01:34
So that's going to cancel out all of the fractions.
01:39
So you can see the denominator of the 6x plus 7 goes away, and so does the denominator of b.
01:47
And then you can see for the denominator underneath a, the x plus 2 will go away and will be left with with an x plus 2 in the numerator.
01:57
So i can rewrite this then as a 6x plus 7 with no denominator equals a times a singular x plus 2 and then plus b because the the denominator for b cancels completely with x plus 2 squared.
02:14
Now we want to solve this equation.
02:17
In order to do that, start by distributing the a into the parentheses.
02:22
So i have 6x plus 7 equals ax plus 2a plus b.
02:28
Now the next step with partial fractions is going to be to group together like terms.
02:34
So i'm going to group together the linear terms, which in this case we just have one of them, and then group together the constant terms.
02:42
Now we can use this to write an equation, because notice that the linear term ax is going to be equivalent to 6x.
02:53
So if ax is equal to 6x, that means that a equals 6.
02:57
6.
02:58
Now notice that for the constant term i have 2a plus b but we know the constant term should equal 7 so i can write the equation 2a plus b equals 7.
03:10
Now i already know that a is going to equal 6 so i can use this information to do substitution for the second equation.
03:21
So plug in a equals 6 and now solve for b.
03:26
So i have 12 plus b equals 7.
03:29
Subtract 12 from both sides to get b equals negative 5.
03:34
So i have that a equals 6 and b is negative 5.
03:38
You can see that's exactly what we got here in the example that's worked out for us...