Integrals Example Number 5 Let’s move on to the next one. This is going to be the integral of 3/x3 + 2x3/2 - 1 dx. All of these problems by the way, the reason why I have parenthesis around them like this, basically the reason that they do that is so that you know that the integral or antiderivative notation which is this thing here and the dx apply to this whole problem and leave them off because I feel like it’s fairly obvious. So, if you have parenthesis like this in you’re book, don’t worry about it. It doesn’t change anything but we’ll go and leave them there for now. The first thing that we want to do is to go ahead and change this first term to a non-fraction so that we can deal with it more easily and the word problem from last time the rule that we had, if you want to move something from the denominator to the numerator or vice versa, you simply change the sign on the exponent. So, this is a positive three, we would change it to a negative three to move it to the top. So, we’re going to say, 1/x3 is the same as x-3, so that’s what we’re going to do. I’m going to change this to be 3x-3 + 2x3/2-1 dx. We can take it turn by turn. We’re going to go ahead and do this term first. We’ll write the x down here and we’re going to add 1 to the exponent. -3+1 is -2 and then as always, divide the quotient on this term, divide by the new exponent, so it’s going to be -3/2. So, there’s the new term there then we want to apply to this term x3/2. We want to add 1 to this exponent. So, 3/2+1 might not be immediately apparent the way that we deal with that. 1, you can change to 2/2 so that the denominators here are the same and you can easily add them, 2/2 is obviously 1. So, we change this to be the same as the denominator here so we can easily add them. So, the result there is 3+2= 5/2. So, adding 1 to this exponent gives us 5/2 and then divide the coefficient here 2 over the new exponent 5/2 and we’ll deal with that later and then minus the antiderivative of 1 is always x as we know and then of course +c for a constant. We have our antiderivative written out but we need to simply some things. We got an integer exponent here. We’ve got 2 over a fraction, so we want to make this simpler. So, what we’re going to do, we shouldn’t have a negative exponent in our answer, so we’re going to use that rule, move this to the bottom and change the sign on the exponent. We’re going to have -3/2x2. We move to the x to the -2 to the denominator and just change the sign on the exponent. So, this term is now simplified. Then we’re going to say +2/5/2. Whenever you have 2 over a fraction like this, the way to deal with this easily is to instead of dividing by a fraction is to multiply by the inverse of that fraction. So, instead of 2/5/2, you can do 2×2/5 and that’s easier to do. You can call this 2/1 then you just multiply the numerators together and the denominators together, so this is going to be 2×2 is 4 over 1×5 is 5, so 2/5. Instead of dividing by a fraction, multiply it by the inverse of that fraction and you get the answer 4/5. We’re going to change this coefficient here to 4/5x5/2–x+c. So, this is our simplified answer here.