This video from IntegralCALC shows you how to solve the Integration by Parts Example2 Part1 Math problem.
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Integration by Parts Example2 Part1 Hey everybody, welcome back, another integration by parts problem. We’re going to go ahead and write the formula at the top so that we can reference it. The formula for all integration by parts problems, integral of udv = uv minus the integral of vdu. Let’s go ahead and write our problem. This one is going to be the integral of ln(x)/x × vxdx. So this formula shows us what we need to do. This part of the formula represents our problem her so we need to assign U and dv two values in our problem. Once we do that we can solve for du and v to get all of these components over and we’ll end up plugging things back in to this formula. The goal being to make this integral over here simpler than the one we start out with originally because to integrate this we don’t really know how we is this method to make this one simpler so that we can end up with an easier integral. So the first thing that we need to do, you may notice ln of x is one term, x is another and the vx is another. So we’ve got three terms in this problem. We only have two terms in our formula here u and dv. So that right there tells us that we need to simplify before we even assign u and dv. So let’s go ahead and do that. This is actually going to become—you remember that the vx is the same as x1/2 they’re the same thing. So this would actually look like the integral of ln of x over x times x1/2. So I just converted the square root of x to x1/2. So now, you can see when you have variables like this multiplied together it’s the same variable that both x’s. You can actually combine then by adding the exponents. There’s an implied one as an exponent on this x so instead of x times x1/2, we can combine this and this actually becomes x and then we add the exponents remember so 1 + 1/2 which is actually 3/2, 1 1/2 or 3/2. So this is x3/2. So let’s go ahead and erase this, I don't get a lot of room, so I'm going to walk you through the steps and then kind of cheat and put this back in here. So we converted this to x3/2 so let’s go ahead and say this is lnx over x3/2. So now, we have two terms, lnx and x3/2. and we’ve got two terms up her so we should be able to assign these values to this two terms, the only thing that I would like to do, I don’t like to integrate with fractions because it’s too confusing for me so I was like to bring whatever’s on the bottom to the top of the fraction if I can. And in this case I can because I only have the one term and the way that I do it is the integral here, I bring this to the top by changing the sign of the exponent, right now it’s positive 3/2, I’m going to make it a negative 3/2 and that allows me to bring it to the top. So x-3/2 and then ln xdx, so now we’ve got two terms, no fraction. So this is something that I’m much more comfortable working with. So let’s go ahead and assign u and dv and the way that we assign u and dv, you always want to or the way I always go about approaching these problems is look for something in here that will— when you take the derivative of that term. It becomes simpler. So what I mean by that is for example, if you assign u here and this, remember x-3/2 is one term, lnx is another term. So if you assign u to x-3/2 and then you take the derivative of that, you’ll end up with -3/2x-5/2 which doesn’t seem a whole lot simpler that just x-3/2. On the other hand if you assign u to lnx, and then you take the derivative, the derivative of lnx is 1/x which does seem simpler to me, the ln or the ln goes away, it’s just 1/x, that’s simpler and in fact if you see lnx in an integration by parts problem, it’s probably got a u because, to simplify this from lnx to 1/x is a good bet for assigning u to the lnx. So let’s go ahead and do that, we’re going to write over here, u = lnx, so we assign u which means dv has to be the other term in the problem, so dv is automatically—and let’s write this down here, dv is automatically x-3/2. So now what we need to do is find du which is the der
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