 Let's look at some examples of computing indefinite integrals using this technique of u substitution, where u substitution is supposed to be the inverse operation for anti-derivatives of the chain rule for differentiation. So when you, the chain rule are this u substitution you often want to use when you see this factorization that shows up here. Like we notice there's this factor of 6x and this factor of 3x squared plus four to the seventh. So factorizations is a big part of recognizing u substitution, because after all you have to identify a function g of u and then this differential d of u. But more importantly we have to recognize composition of functions. We want to recognize a function inside of a function. So notice in this example we have this 3x squared plus four, which sits inside of the seventh power function. So when we see this function inside of a function that's often the sort of the thing that triggers this whole u substitution process. It's the catalyst to it. If you have some superpowers maybe your spider sense starts tingling like oh u substitution is coming because you see this function inside of a function. And so I'm very inclined to suggest that you should equal 3x squared plus four for this example right here, because we have this function inside of that one. Now in order for u substitution to work its derivative has to also be present, because I like to think of anti differentiation as like plain detective like I'm a detective and I come to the scene of the crime it's like I see what's happened 6x times 3x squared plus four to the seventh. It's the derivative of but of who right who took the derivative who done it. And so it's we round up the usual suspects the power rule, the chain rule the product rule from differentiation right who done it right and we're looking for the evidence here. And now these these master criminals, they like to mock the police officer like to mock the detective so they leave their calling cards behind. And for you substitution right that is the chain rule, if it's the one who took the derivative, then it always leads its calling card behind the calling card is that inner derivative, we have to see if the inner derivative is there. So if we think the original function was 3x squared plus four, then taking the inner derivative, we would end up with a 6x dx. And we can see that those pieces are exactly here in the integral, we always have the dx the differential is always to be there. It's the 6x part that we need, putting these together, we form our du. And so therefore we can calculate the anti derivative using this u substitution. So plugging in the substitution here, the we're going to take the integral of 3x squared plus four to the seventh the 3x squared plus four becomes our u. So we get you to the seventh. And then the 6x times dx all comes together to give us a du that all comes together, the six and the x aren't really disappearing we're just substituting something equal for them. Instead, so the 6x dx because our du by the anti power rule. We raise the power to the eighth, we divide by eight at an arbitrary constant, and then substitute in the original value of you in terms of x here. So we end up with one eighth 3x squared plus four raised to the eighth power plus a constant. And this gives us our anti derivative. If we want to check if we have the right correct anti derivative we can take the derivative of this thing. I'm not going to do that right here but I encourage you to pause this video to check it yourself just to verify your own derivative techniques here, but we are able to find this anti derivative here using this u substitution. Let's look at our next crime scene detectives. We have, we see in front of us the integral of x squared times the square root of x cube plus one. Well, the fact we have a x cube plus one inside of a square root makes me start to believe well maybe you substitution is that play right here right because we have this x cube plus one, which sits inside of the square root function right here we have this chain going on. And so I'm inclined to believe I could do a u substitution using that inner function x cube plus one. But if I'm going to, if I'm going to do this inner the sorry if I if I'm going to do the u substitution we have to have the inner derivative, which is going to be three x squared dx, which well let's see what we have right here. We have a dx. That's good we have an x squared. But what about a three. I don't have a three anywhere. What are we going to do without a three. Does that mean that's not the u substitution that we're going to use right here what can we do what can we do. Well, as you sit there pondering for that for a moment look out a window maybe oh is that a cute little butterfly flying through the window you're not looking right you're not looking right you're not looking right oh oh come back to the screen everyone. Oh look we did have a three there the whole time right we have a three x squared dx inside. What's going on here is that if you're missing your constant multiple like we're just missing the three. If you're missing the constant multiple that's easy to fix. We can actually put a three inside of the integral just by dividing by three as well. Three over three is equal to one and the strategic number one helps us out here. So that when we put these things together now we get three x squared dx. This comes together and makes Captain Planet know these powers combined form du instead. And so by this technique of you substitution we get one third the integral of the square root of you du are in terms of power functions you might prefer you to the one half du. So if you're only off by a scalar multiple you can always correct that by times it and dividing by the scalar multiple you want to make the inner derivative correct here and just make sure you keep track of this one third. As the price you pay for not having the three in there in the first place. And then doing the power rule for anti derivatives this can get confusing sometimes especially with you substitution. Because with you substitution you have to take the derivative to find the inner derivative then you want to anti differentiate because you'll be taking derivatives and anti derivatives and same problem sometimes we get mixed up which way we're doing. So make sure that you use the anti derivative version of the power rule. And so you're going to get one third times raise the power of one half to be three halves so one half plus one is three halves. Then you divide by that new power three halves, add your arbitrary constant. Now of course if you're dividing by three halves that's the same thing as times in by two thirds. So you're going to get a coefficient of two nights in front. If we replace you with its original expression in terms of x, xq plus one, we then get two thirds xq plus one to the three halves power, plus an arbitrary constant, which gives us our anti derivative. This is the general anti derivative of this function right here. Now in that last example we saw that if we were missing the the scalar multiple in front of the inner derivative we are able to fix that by times in like in that case three over three. So how much wiggle room do we have when it comes to I don't have the right anti derivative. Honestly not a lot basically that scalar multiple is the only flexibility we have for this technique of you substitution and I want to show you an example. What can go wrong right. If you look at this function right here, you want to find its anti derivative x cube times the square root of x cube plus one DX here. It's very similar except the out of the function outside the square root is an x cube instead of an x squared. So we might be tempted to try to make the same you substitution work right. We have a you equals x cube plus one, and so then do you would still be three x squared DX. Like so, and that's not exactly what we have right here what we have is we have an x times the square root of you let's make that transition right now. And then we have this x squared DX right here. And so we could stick a three in front of that by dividing by one third and one front. And so making that substitution we end up with this one third, the integral of x, you to the one half du. And so we've kind of partially substituted the new variable you instead of x so how do we deal with the x right there it's like oh no there's there's there's not a whole lot you can do with that. Later on in calculus to we will kind of talk about some things one could do but it turns out substitution doesn't work very well on this because you're going to get this extra bit this x that we really can't do much about if you're missing a multiple that's fine. But if there's more to it than just a skill multiple see you substitution isn't going to be very effective here. This one actually you'd want to use integration by parts, which is a technique that we can see later in calculus to.