 Alright, let's take a look at some more anti-derivatives. So let's find the anti-derivative of quantity x cubed plus 4x squared, where our differentiation is taken with respect to x. Notice that when we indicate the anti-derivative, we do have a parentheses around the function whose anti-derivative we're looking for. If you leave off those parentheses, it becomes a statement that doesn't make any sense mathematically. So those parentheses are very important. So first thing to note here, again, a little bit of analysis goes a long way. I'm anti-differentiating a sum of two terms, and I think back through my derivative rules, the derivative of a sum is the sum of the two derivatives. And so what that suggests is that if I want to find the anti-derivative of a sum, I can find the sum of the anti-derivatives of the two add-ins. So here I'm adding x to the third, 4x squared. And so if I find the anti-derivative of x to the third and the anti-derivative of 4x squared, then I should find the anti-derivative of the sum of the two functions. So let's start off with x to the third. So x to the third is an x to the n-type function. So I might start off by looking at the derivative of x to the fourth, which gives me 4x to the third, close, but not quite. And if I want to get closer, I'll multiply by 1 fourth on both sides. That gets me my x cubed, and I have constant times derivative is the same as derivative of constant times function. So anti-derivative of x cubed is 1 fourth x to the fourth. Likewise, I'm looking for a function whose derivative is 4x squared, and that's constant times function, and the type of function is an x to the n function. So again, I might start with an x to the n plus 1. So if I want a derivative x squared, I'll start with derivative of x cubed, and that gives me derivative 3x squared. Not quite what I want, but I can modify it by multiplying by 4 thirds both sides, and that gets me over on the right hand side what I want as the derivative, and over on the left hand side, constant times derivative of function, which is the same as derivative of constant times function. So again, what do I have? Well the derivative of 1 quarter x cubed is x cubed is the first term, the derivative of 4 thirds x cubed is 4x squared is the second term, and the derivative of a sum of two functions is the sum of the two derivatives. So if I add those two functions together and differentiate them, I'm going to get x cubed plus 4x squared. So that tells me this function here, whose derivative is x cubed plus 4x squared, is going to be an anti-derivative. So the anti-derivative is this function plus any constant value. Well how about some other functions? So for example, the anti-derivative of sine of x, and so some functions you just have to recognize. So here I am looking for a function whose derivative is a trigonometric function, and I might not remember which one that is, but there aren't that many of them. So if I think about that, the trigonometric functions all have trigonometric derivatives. If I can't remember which one gives me sine of x, I can just find all of them. So what is it? Derivative of sine is cosine, not what I'm looking for. Derivative of tangent is secant squared, again not what I'm looking for. Derivative of secant is a horrible mess, again not what I'm looking for. Derivative of cosine is almost what I'm looking for. Derivative of cosine is minus sine of x, and almost counts in calculus. So let's see, I want to modify that. I want sine of x over on the right hand side, so I'll multiply both sides by negative one. And now I have constant times derivative of function, so I can move that constant inside the differentiation operator. And so the derivative of minus cosine x is sine of x, and so my anti-derivative minus cosine x plus any constant C that we want. Well how about the anti-derivative of one over x? And I can do a little bit of algebra, one over x is the same as x to power negative one. And I know what to do with an x to the n, I'm going to differentiate an x to the n plus one. So negative one plus one is zero, so the derivative of x to the zero gives me zero, x to power negative one, and there's my x to the negative one, and all I have to do is divide by zero to get rid of this coefficient. Ah, except I can't divide by zero. So we have a problem with this one, and unfortunately this is one of those anti-derivatives that requires you to actually know something. And in this particular case what you have to know is the derivative of log is one over x. So remember what this is asking for. You want to find a function whose derivative is one over x. So thinking about it, you know a function whose derivative is one over x. So the anti-derivative of one over x is going to be log of x plus any constant that you want.