 Let us evaluate the integral of the square root of one plus e to the x. And I wanna do this using a ratio sub. You know, without using the rationalizing sub, the other approach here would be like integration by hope. You don't know what the anti-derivative square root of one plus e to the x is, but I could take its derivative and I just hope there's enough similarity that I can make something work there. So instead of putting all of our hope into hope, integration by hope here, we can actually use the rationalizing substitution, for which case we're gonna take u to be the square root of one plus e to the x. The reason we do that is because when we have a square root or any kind of radical in fact, there's a similarity between the function and its derivative and we can manipulate that similarity to help us get something easier to integrate. So by our usual rules here, by the chain rule, when you take the square root's derivative, you get a two times that exact same square root on the bottom. Then the inner derivative goes on top, which here would be e to the x dx, like so. Now we know what to do with the square root of one plus e to the x, that itself is equal to a u. And if we manipulate the original equation, right, square both sides, you get u squared is equal to one plus e to the x, subtract one, e to the x is equal to u squared minus one. And so that's what we're gonna do here. Coming back, du is gonna equal u squared minus one, dx over two u, solving for dx, we get dx equals two u du over u squared minus one. So making that substitution, remember the square root of one plus e to the x, that is a u, dx is equal to two u du over u squared minus one, like so, right? And therefore, with this now, we can try to simplify this thing. And when I say simplifying, the fact that we have a u squared over u squared minus one, I'm gonna try to, because this is a proper rational function, I'm gonna try to squish it down, u squared over u squared minus one, I'm gonna do it over here as well. I want a u squared minus one on the top. In order to get a minus one, I have to add one here. This will just reduce down to be a one, the anti-derivative of one with respect to u, of course, will just be u, so we get two u. And then we get the integral of two over u squared minus one du. This one right here, two over u squared minus one, we could try to do sometimes a secant substitution, but I think partial fractions will be much cleaner in this situation, because our partial fractions will look like a over u minus one, and then the integral of b over u plus one, just factoring u squared minus one there. And if we proceeded to do that, we're gonna get two times a u plus one plus b u minus one using the partial fractions there, the technique. We would annihilate first using say like one, that'll kill off the b, in which case we get two is equal to two a, equals one. If you annihilate using negative one, that would annihilate a, and so you'd get two equals negative two b, in which case we then see that b equals negative one, like so. And so putting that in above, we get two u, a, we got us one, so we're gonna get the natural log of u minus one, b was negative one, so we get minus the natural log of u plus one, plus a constant. I can combine those logarithms together, because you're taking a difference of logs there. That would look like, oh, there's supposed to be u right there. So two u plus the natural log of the absolute value of u minus one over u plus one, plus a constant. And now remember what u was, you remember was the square root of one plus e to the x. So this becomes two times the square root of one plus e to the x, plus the natural log of the square root of one plus, e to the x minus one, over the square root of one plus e to the x plus one, plus a constant, like so. And so again, the rationalizing substitution saves the day. This is the, I would say the cleanest technique to calculate this one, but you could have tried something like integration by hope or something like that.