 Hello, and welcome to yet another screencast about integration by substitution. Hopefully by now you guys are kind of getting the swing of things. So I'm going to start this one out a little bit differently. Okay, we're going to evaluate the indefinite integral, and this is 3 divided by 2x plus 3 dx. All right, instead of me telling you right away what we're going to integrate, I've got some multiple choice questions here. So I want you to pause the video, look over your four choices, pick one, and then decide, and then go from there. Okay, so you can either work through the whole problem to make sure that that's the right choice for you, or you can obviously start the video again. Okay, so if you go through your choices, definitely B is going to be the best one. Okay, now let's discuss some of these other ones and see kind of why they will or will not work. So 2x, I mean that's a reasonable idea for you substitution. But the problem is, is really the inside function with this one is your whole denominator. If you only let you be that first piece of your denominator, then you've still got this plus 3 to deal with, and then what do you do with that? Okay, so that's probably not the best choice for sure. Okay, so I crossed that one out. 3, constants are never good choices because what are their derivatives? Zero, yeah, don't want that. And then using the whole function is probably not a good idea either, because remember, you want to pick your U substitution, so your integral is easier, not the exact same thing. Okay, so that's not going to work. Because if you have to do the derivative of this function, you're either going to have to rewrite it and use a chain rule, or oh my gosh, I might have to use a quotient rule, or you know, it's going to be ugly. Okay, so anyway, using this particular substitution would not make it easier. Okay, so I totally agree that the best choice here for our substitution is picking that denominator. All right, so let's write all this out then. So if we let U be 2x plus 3, my du then is 2dx. All right, now here goes our matching game again. So now we have to figure out what we need to do. Does this 2 make sense? Do we need to do some other stuff to it? So let's look back at our original integral here. So remember the 3, that's a constant. I can just pull that out front. So really, what I'm trying to get then is just that dx by itself. Okay, and that's typically what you're going to want to do with most of these problems. So how do we get rid of that 2? You can either divide by 2 or multiply by a half. I like to multiply by a half, because I think it looks a little bit nicer. So that's going to be my dx. All right, so then, matching everything up then with my original integral, I've got the integral of 3, because that didn't go away, divided by U, and then my dx is 1 half U, or du, sorry. Okay, well, this is really ugly, so let's do some finagling with this one. If we bring the 3 out front, we can do that. We can also bring this 1 half out front, so that gives us a 3 halves for our constant. Then we're going to end up with the integral of du over U. Okay, because remember this is technically du over 1, so we're multiplying fractions in here if you want to. If you also want to write this as 1 over U du, that's also the same thing. So totally up to you how you want to write it. Okay, so now we've got to remind ourselves, hmm, what function when I do the derivative of it will give me 1 over U? Well, hopefully you've got this one in your arsenal, so that is the natural log. And it's always the absolute value of U. And then again, this is an indefinite integral, so we want our plus C. Okay, so my 3 halves again just came along for the ride. My integral or my antiderivative of 1 over U is natural log of U, and then I went ahead and put my plus C on. Okay, let's plug in our U, and then we are done. So the natural log, or sorry, 3 halves natural log, absolute value of 2x plus 3 plus C is our final answer. And again, to check and make sure this is right, I'll leave that up to you, is you want to take the derivative and make sure that you get back to exactly where you started. Great, thank you for watching.