00:01
For this problem, we are to find the integral using partial fractions.
00:04
Now, we begin by finding the partial fraction decomposition for the expression 4t squared plus 3t minus 1 over t -cube minus d squared.
00:17
Now note that 4t squared plus 3t minus 1 over t to the third power minus t squared can be written as 4t squared plus 3t minus 1.
00:30
Over t to the second power times t minus 1.
00:34
Now since we have a repeated factor in our denominator, then the partial fractions will have denominators in powers of t.
00:45
So we have a denominator of t and a denominator of t squared.
00:52
And then lastly, the linear factor, t minus 1.
00:57
Now since all of these are linear, then their numerators are constants, that say, a, b, and c.
01:06
Now from here, we want to solve for the values of a, b, and c by first multiplying this by the lcd t to the second power times t minus 1.
01:20
So from here we have 4t squared plus 3t minus 1.
01:25
This is equal to a times t times t minus 1 plus b times t minus 1 plus c times t squared.
01:38
Next we want to fix values for t to be used in this equation so that we can solve for a, b, and c.
01:46
Now we can choose any t values, but for convenience we will choose those t values where the denominators are zero.
01:56
In here, we will first set t equals 0, and we will use this to find a, b, or c in this equation.
02:06
So when t is 0, this means that 4 times the square of 0 plus 3 times 0 minus 1, this is equal to 0 plus b times 0 minus 1 plus c times 0, that's 0...