00:01
So i have an approach on how i would do this integral problem.
00:09
What i would do is let u equal x plus 3.
00:15
So then du is going to equal the derivative.
00:19
Well, the derivative of x is just dx, and the derivative of 3 is zero.
00:24
So let me rewrite this real quick.
00:27
It's going to be equal to an integral, but my bounds need to change.
00:31
Let me just write it like this, x over u squared, because remember i let, this x plus 3 squared, sorry, x plus 3, i let that equal u, so it's still squared, and this dx i can replace with du.
00:46
But we have two issues.
00:48
One issue is we have x's with u's, and we also have to change these bounds.
00:53
Let me do that real quick.
00:54
If i plug in negative 1 for this x, negative 1 plus 3 would give me 2, and then if i plug in 1 in for this x, 1 plus 3 gives me 4, we have to replace those u's.
01:06
But we also cannot have this x here.
01:09
So here's my solution to that.
01:11
As i can rewrite this problem as subtracting this 3 to the other side.
01:16
So x is going to equal u minus 3.
01:19
So let me substitute that into this problem.
01:25
So i'm going to replace this x with what it's equal to, which is u minus 3.
01:29
But then i'm also going to, instead of write it as division, divided by u squared, i can rewrite that as u to the negative second power...