We already know. When we differentiate log X. We end up with. Is one over X? We also know that if we've got Y equals log of a function of X. And we differentiate it. Then what we end up with is the derivative of that function over the function of X. Now the point about integrating is if we can recognize something that's a differential, then we can simply reverse the process. So what we're going to be looking for or looking at in this case, is functions that look like this that require integration, so we can go back from there to there. So let's see if we can just write that little bit down again and then have a look at some examples. So no, that is why is the log of a function of X then divide by The X is the derivative of that function divided by the function. So therefore, if we can recognize. That form. And we want to integrate it. Then we can claim straight away that this is the log of the function of X plus. Of course a constant of integration because there are no limits here. So we're going to be looking for this. We're going to be looking at what we've been given to integrate and can we spot? A derivative. Over the function, or something approaching a derivative. So now we've got the result. Let's look at some examples. So we take the integral of tan XDX. Now it doesn't look much like one of the examples. We've just been talking about, but we know that we can redefine Tan X sign X over cause X. And now when we look at the derivative of cars is minus sign so. The numerator is very nearly the derivative of the denominator, so let's make it so. Let's put in minus sign X. Now having putting the minus sign, we've achieved what we want. The numerator is the derivative of the denominator, the top is the derivative of the bottom. But we need to put in that balancing minus sign so that we can retain the equality of these two expressions. Having done that, we can now write this down. Minus and it's that minus sign. The log of caused X plus a constant of integration, see. We're subtracting a log, which means we're dividing by what's within the log. Function, so we're dividing by cause what we do know is with. With dividing by cars, then that's one over cars and that sank. So this is log of sex X Plus C. Now let's go on again and have a look at another example. Integral of X over one plus X squared DX. Look at the bottom and differentiate it. Its derivative is 2X only got X on top, that's no problem. Let's make it 2X on top by multiplying by two. If we've multiplied by two, we've got to divide by two, and that means we want a half of that result there. So now this is balanced out and it's the same as that. What we've got on the top now is very definitely the derivative of what's on the bottom, so again, we can have a half the log of one plus X squared plus C. We can even have this with look like very complicated functions, so let's have a look at one over X Times the natural log of X. Doesn't look like what we've got does it? But let's remember that the derivative of log X is one over X. So if I write this a little bit differently, one over X divided by log X DX. Then we can see that what's on top is indeed the derivative of what's on the bottom, and so, again, this is the log of. Log of X plus a constant of integration. See, so even in something like that we can find what it is we're actually looking for. Let's take one more example. A little bit contrived, but it does show you how you need to work and look to see. If what you've got on the. Numerator is in fact the derivative of the denominator. So let's have a look at this. Looks quite fearsome as it's written, but let's just think about what we would get if we differentiate it X Sign X. I'll just do that over here. Let's say Y equals X sign X. Now this is a product, it's a U times by AV, so we know that if Y equals UV when we do the differentiation. Why by DX is UDV by the X plus VDU by DX? So in this case you is X&V is synex, so that's X cause X Plus V which is synex. Times by du by DX, but you was X or do you buy X is just one? So if we look what we can see here. Is that the numerator? Is X cause X sign X, which is the derivative of the denominator X sign X, and so again complicated though it looks we've been able to spot that the numerator is again the derivative of the denominator, and so we can say straight away that the result of this integral is the log of the denominator X Sign X. Sometimes you have to look very closely and let's just remember if we just look back at this one. But sometimes you might have to balance the function in order to be able to make it look like you want it to look. But quite often it's fairly clear that that's what you have to do. So do remember, this is a very typical standard form of integration of very important one, and one that occurs a great deal when looking at differential equations.