00:01
To integrate this problem, we're going to set this up as two separate integrations.
00:13
And we're going to bring each denominator up to the negative one power.
00:25
So each problem will be a u substitution.
00:28
So the first one, u is 2x plus 5.
00:32
The du then is 2dx.
00:35
And the second one, u is 2x minus 5.
00:39
And the du then is 2dx.
00:45
So we are integrating u to the negative 1.
00:47
And we have to match up with the du on the first one.
00:51
We have the dx.
00:53
We need the two, and you need to put a one -half out front.
00:58
In the second integration, we need again to match up with the du.
01:06
We have the dx, we need the two, so we compensate and put a one -half out front...