 Let's take a look at how to evaluate some anti-derivatives. Remembering the power rule for integration, we're going to have to take each variable quantity, add one to the exponent, and then divide by that new exponent. And we're going to work one term at a time. We can bring the five out in front for the first term. Now we're doing the anti-derivative of x to the fourth. So we need to add one to the exponent, and then we divide by that new exponent. Now we move to the second term. We'll keep the eight as the coefficient. We have x to the fourth as we add one to the exponent, and then we divide by that new exponent. The nine remains as our coefficient in the third term. Add one to the exponent, divide by the new exponent. Now in the fourth term, remember, of course, the exponent is assumed to be one. We keep the two as the coefficient, and we multiply that by x squared over two. The last term is assumed to be x to the zero power. So as you add one to the exponent, it becomes x to the first. You could also think of it considering the fact that derivatives and anti-derivatives are inverses of each other. If you were to take the derivative of 7x, you get 7. So you could think of it as, what do I have to take the derivative of to get 7 as my answer? So that's really where 7x comes from. You can think of it as two different ways. Then of course, we need plus c, because as you saw in the introductory activity, there could have been a constant term on this original function that when you took the derivative of it, it became zero, and zeroed out. So you need your plus c, your constant of integration, to account for the fact that there could have been a constant term affixed with this function. Now we just simplify. We end up with x to the fifth minus 2x to the fourth plus 3x cubed minus x squared plus 7x plus c. One of the most wonderful things about anti-derivatives and what most people really like about them, you have a way to check your answer. Now that we've arrived at this as our anti-derivative, take the derivative of this. You should get your integrand back again. If you don't, you made a mistake. That's something that we never had with derivatives, a way to check our answer. Here we do. So please do take advantage of that. Let's try another one. Now as we go through these, if you think you're getting pretty good at them, please feel free to pause this video, try it on your own, then restart it to see how you did. This is a problem in which we're going to have to algebraically manipulate it. We're going to have to rewrite the square root of x as x to the one-half and distribute it across the quantity x plus 1 over x. So if we distribute, we get x to the three-halves plus x to the negative one-half because remember 1 over x would be x to the negative one, dx. Remember, we're going to add one to our exponent, then divide by that new exponent. Some people find it helpful as they're just starting this out to actually write that out. You get very good at working with fractions in your head. If we add one to the exponent of three-halves, we get five-halves. Then we have to divide by five-halves. But if you're dividing by five-halves, remember that's the same as multiplying by the reciprocal two-fifths. So we have two-fifths x to the five-halves plus adding one onto the exponent of negative one-half, we get a positive a half, but then we're dividing by it, so we're multiplying by two. And then, of course, you need your constant of integration. Again, take your derivative of this. You'll notice you get this integrand of x to the three-halves plus x to the negative one-half back again. Yet another one we'll have to simplify using our laws of exponents. If we have t squared over t to the four-thirds, subtracting the exponents, we get five t to the two-thirds, and then we have seven t to the negative four-thirds. We're going to add one to our exponents. So the five, this five here, that's going to remain as our coefficient. If we add one, the three-thirds, onto two-thirds, we get five-thirds, then we have to divide by it, so we're multiplying by three-fifths. Now our seven remains as our constant, doing the same with our exponents. We have t to the negative one-third, but then we're dividing by negative one-third, so that's multiplying by a negative three plus c for our constant of integration. Simplifying this, we have three t to the five-thirds, minus twenty-one t to the negative one-third plus c. Again, feel free to check your answer. I highly suggest that by taking the derivative of your answer and making sure you got your integrand. In our next example, we're doing the anti-derivative of two over x. Now this is one for which, remember, that would be two x to the negative one, the power rule does not work. This is the first of the special rules we looked at. The two will remain as our coefficient. Anti-derivative of one over x, remember, is natural log absolute value of x. Yes, it has to be absolute value. Again, we will talk about y in a later lesson. Plus c. Again, you can check your answer. Derivative of natural log of x is one over x, so you do get two over x back again when you check your answer. Now another way to write this, if you think of your laws of logarithms, remember that two could pop back up to become the exponent of x. So another way you could write this would be the natural log of x squared plus c. When you square that value, that is the absolute value. Of course, it's always going to be positive anyway. So that's why you can lose the absolute value bar. So this would be another way you could express your answer. And finally, our last example. We're doing anti-derivative of 8 to the x. This was another one of the special rules for you to memorize. It's going to be 8 to the x divided by natural log of 8 plus c. Let's go ahead and check this one to make sure it's right. So if we took the derivative of this, remember one over natural log of 8 remains because it's a numerical quantity. Now we're just doing derivative of 8 to the x, which would be 8 to the x, natural log of 8. The derivative of that constant, of course, would just be 0. Notice how your natural logs cancel and all you're left with is your 8 to the x.