00:01
Consider the integral of y squared minus y plus 2 over y cubed minus 1 dy.
00:06
To evaluate this integral, first thing you have to do is to rewrite the integrand as a sum of its partial fractions.
00:14
Now y squared minus y plus 2 over y cubed minus 1 can be rewritten into y squared minus y plus 2 over we have y minus 1 times y squared plus y plus 1.
00:32
So then we can write this as a sum of partial fractions with denominators y minus 1 and y squared plus y plus 1.
00:43
So for the denominator y minus 1, since it's linear, it will have a constant numerator, let's call it a, and for the quadratic denominator y squared plus y plus 1, it will have a linear numerator, let's call it b, y plus c.
01:00
So to find a, b, and c, first thing you have to do is to multiply both sides of the equation by the lcd y minus 1 times y squared plus y plus 1.
01:13
So then from here, we'll get y squared minus y plus 2 equal to a times y squared plus y plus 1 plus we have b y plus c times y minus 1.
01:28
Expanding that, we have y squared minus y plus 2 equal to a y squared plus a y plus a plus b y squared plus or minus b y plus c y minus c.
01:47
And then that'll be y squared minus y plus 2 equal to a plus b times y squared plus you have a minus b plus c times y and then plus a minus c.
02:03
So comparing now the coefficients of the variables at the left side of the equation and at the right side, we can form a system of equations.
02:15
It'll be a plus b equal to 1, a minus b plus c equal to negative 1, and a minus c equals 2.
02:27
Now adding the first two equations, we will have 2a plus c equal to 0.
02:38
And then adding the third equation from this equation, so adding a minus c equal to 2, we will have 3a equal to 2, meaning a equals 2 over 3...