00:01
For this, rob, we want to find the indefinite integral of t over 1 minus t cubed dt using power series.
00:09
Now, we recall 1 over 1 minus x equals the summation from n equals 0 to infinity of x raised to n.
00:16
So this we can rewrite into the integral of t times 1 over 1 minus t to the third power d t.
00:25
And then using this, we have integral of t times the summation from n equals zero to infinity of t to the third power raise a power n d t.
00:40
That's integral of the summation from n equals zero to infinity of t raised the 3n plus 1 d t.
00:51
Now expanding this, we have the integral of when n is 0, that's t raised a power of 1 or t, plus when n is 1, we have 3 plus 1 or 4, plus when n is 2, we have t raised to 7.
01:10
When n is 3, we have t raised 10 and so on, and then d t.
01:19
So integrating term by term, we have t squared over 2 plus t raised to 4 plus 1 or 5 over 5 plus t to the 8th over 8 plus t to the 11 over 11 and so on.
01:44
That's the same as, oh, by the way, we have a plus c here since we're integrating and we have a constant of integration.
01:53
And this is equal to the summation from n equals 0 to infinity of t raised to 3n plus 2 over 3n plus 2 and then plus c.
02:08
We could get this pattern from the original form here, which is 3n plus 1n.
02:14
Remember in power rule, when we take the antiderivative of, let's say, x raise a power of n, this is x raise 2.
02:23
N plus 1 over n plus 1 so that 3n plus 1 is added by 1 and we divide the same form of the exponent to the denominator it becomes over 3n plus 2 so this is our integral and for the radius of convergence we'll have to apply the ratio test now in this test we want the limit of the absolute value of a sub n plus 1 over a sub n as an approaches infinity to be less less than 1...