00:01
6 .7, proven of number 5, integration of my parts, the integral of q to the 5th, natural log of 5q.
00:09
Keeping in mind the driving force behind this entire section is that the integral of udv can be written as uv minus the integral of vdu.
00:21
This is sometimes a way to simplify when i have two functions multiply.
00:26
So the key is you want to try to pick a lot of times you being the polynomial, the exception, when i see a natural log, it's quite often easier to let you be that log term, because when you take the derivative of the log, that simplifies the natural log goes away.
00:41
So if u is equal to the natural log of 5q, then du is going to be 1 over 5q times 5dq.
00:58
So that is just going to say that du is equal to 1 over q, dq.
01:07
And then dv, if you look at this integran, what's left is q to the 5th, dq, which means v is equal to, when i integrate that, i get q to the 6 over 6.
01:22
So this integral that you see here can be written as uv, so that is going to be q to the 6 to the 6...