 In this last video for section 16, the street fighting integrals here, I wanna do one more example of using a rationalizing sub to help us out here. We've seen a couple of examples for which because there's square roots or other radicals involved, you could do a ratio sub, setting U equal to the radical, square roots is the one we saw the most often. And you use the self-similarity between the square root and its derivative to work out a nice substitution here. The square root is not the only function which is self-similar to its derivative. In this situation, notice, e to the x is its own derivative. And if we used that here, we could take sorry, U to equal, equal e to the x, and hence du equals e to the x dx, which is to say that dx equals du over u. And so what we could do then is we could then get an integral that looks like, well, the dx, we'll do the first one first. We get one over one plus u times dx, which becomes du over u. We could do some type of partial fraction decomposition here for which case your integral, we're looking for a over one plus u du plus the integral of b du over u like so. In which case, let's do the calculation for that. So we have one is equal to a u plus b times one plus u. Kind of already took the liberty of clearing the dom there. So the template of our partial fraction decomposition, we should plug in u equals zero. That would annihilate the a. And so then we get that b equals one. If you stick in u equals one, that not one, sorry, that's not one. Sorry, negative one. That annihilates the b. And so then we see that negative a equals one. So a equals negative one like so. So we end up with negative u plus one plus u. That works out just great. So coming back to this part right here, bring that down. We have the integral of negative one over one plus u du. And then we have the integral of one over u du. We then get negative the natural log of the absolute value of one plus u plus the natural log of the absolute value u plus a constant. We could put those things together. We would get the natural log of the absolute value of u over one plus u like so plus a constant. And then remember that u is itself equal to e. We get the natural log of the absolute value of e to the x over one plus e to the x plus an arbitrary constant. And so this is an acceptable way of writing the final answer right here. I also wanna mention that since we're taking the natural log of e to the x, had you not combined the logs together, if you had kept it looking something like this, you're gonna get the natural log of e to the x. So that would give you an x minus the natural log of the absolute value of one plus e to the x. One plus e to the x is always positive. So if you forgot the absolute value, you should still be correct. So that's an alternative that you could do. That's perfectly acceptable. And so I wanna give you this example of rationalize and substitutions that didn't involve any square roots at all. It's the self-similarity of the function with this derivative that we're using here. e to the x is its own derivative. The derivative of the square root involves that same square and that's what we're able to get away with here. I also wanna make mention that we did the u substitution, u equals e to the x. You could also have gotten away with u equals one plus e to the x because it's derivative. It's still gonna be e to the x dx. And so you have to modify things a little bit here because you're gonna have to use the fact that e to the x equals u minus one. And so that changes things a little bit in which case you can then plug that into here. dx equals du over u minus one. You get a slightly different rational function in that situation. Your integral is gonna look like one over u times one over u minus one du. It's a little bit different but in the end you get the same answer when you're done. So again, there are options you could take when you're doing calculus problems. There's not exactly one right way. There are some ways that are much easier than others and that's what this section 7.5 is all about. How do we find the best technique or in the presence of options, what do we do? What do we do to try to make things a little bit better for us, all right? So that'll actually conclude our lecture 16 about street fighting integrals. I hope you enjoyed it. Next time we're gonna talk about, well, when street fighting doesn't work, it turns out there are some other options and we will see it then. Stay tuned and see you next time everyone, bye.