 Alright, let's see some more examples of calculating anti-derivatives using the power rule plus also the linearity properties That is constant multiples can be taken out of integrals and we can take sums and integrals a sums and difference of integrals here So when you look at something like this the integral of x squared minus one quantity squared If this were a derivative problem, we would be tempted to use the chain rule But for anti-derivatives, we don't have a chain rule yet There is no chain rule for us to be using and as such we can't use a chain rule We might learn something like the chain rule later. It's a little bit more complicated But we'll see this at the end of the semester. It's called a u substitution in the meanwhile though We can utilize some algebraic Simplification because after all if it's x squared minus one square that just means x squared minus one times x squared minus one And if we foil that thing We end up with x to the fourth minus 2x squared plus one and We can take the anti-derivative of that polynomial there by the power rule. We're gonna get x to the fifth over five minus 2x cubed over three and then plus x plus a constant So you'll notice in this example I was able to do each individual term one by one by one and that gives us our anti-derivative X to the fifth over five minus 2x 2 3rds x cubed Plus x plus c All right, the next one here if we want to find the anti-derivative of nine e to the t Well, I would first take out the constant So we have to find nine times the integral of e to the t dt And so now we're like well We have to find a function whose whose derivative is e to the t You probably don't have to search very far for that because we know if you take The function e to the t you take its derivative with respect to t that would just be e to the t again Since so e to the x the natural exponential is its own derivative, which also makes it its own Anti-derivative this is pretty cool. You're gonna get nine e to the t But you have to be careful you have to remember to take this plus a constant right so we get nine e to the t plus a constant and So every every derivative rule that we know can be translated into an anti-derivative rule So the derivative of e to the t is equal to e to the t This is the same thing as saying the integral of e to the x dx is equal to e to the x plus a constant This one is worth saving for a future date there Well, how about the function the integral of 4 over x dx well, I can take that constant out 4 times the integral of 1 over x dx and Note here that the function 1 over x could be written as x negative 1 But you can't use the power rule function for the power rule for anti-derivatives here because the power rule doesn't work When x equals negative 1 we need a function whose derivative is equal to 1 over x and we know a function who does that We know that if you take the natural log of x its derivative is 1 over x Now that's almost right but in terms of domain the domain of 1 over x is everything except for 0 but the domain for the natural log is Only positive numbers if we add in the absolute value of x Then the derivative of the natural log of absolute value of x is still 1 over x And that's actually how we're going to record this anti-derivative the integral of 1 over x dx is Equal to the natural log of the absolute value of x plus a constant and so you're going to want to save this for later Every derivative rule we have can be turned around and become an anti-derivative rule So we see here that the anti-derivative of 4 over x is 4 times the natural log of the absolute value of x plus a constant Don't forget the plus C and also don't forget the absolute values Like so Alright, let's look at another one last example of this Because there's two parts. Let's break up the integral into those two ports We get four times the integral of sign of x dx that's the first indefinite integral and then the next one we're going to take the integral of Well, I'm going to break up that fraction as well because I don't have a quotient rule I can break it up into two different pieces. We get 2x to the fifth over x dx and we're going to get minus the integral of x to the one half Over x dx now. I know exactly what to do on those second pieces. I can simplify those fractions Down so I get a power function. So I get two times the integral of x to the fourth Five minus one there and then for the other one I'm going to get the integral negative integral of I have the one half power and the first power So their difference will give me x to the negative one half dx I can apply the power rule for them, but what do you do for sign? Well, we have to find a function whose derivative is equal to sign and we're kind of close with that We know a function that almost does that It is true that cosine of x its derivative is negative sign of x And we do also know that the derivative of sign is equal to cosine So if we turn those around we get the following We know that the integral of cosine of x is going to equal whoops almost forgot my differential there We know that the antiderivative of cosine of x is going to equal sine of x plus a constant Because the derivative of sine is cosine so the antiderivative cosine is negative sign but for for sine there The integral of sine dx we have to we have to recognize that the derivative of Sinus cosine, but the antiderivative of sine is going to be negative cosine of x plus a cos plus c there and you can double check there if you take the derivative of negative sine The derivative of cos sorry the negative cosine the derivative of the cosine is negative sine So the double negative makes it a positive And so that those those principles we need to remember right here for later Derivative rules turn into antiderivative rules All right, so let's finish this problem here The antiderivative of sine is going to be negative cosine of x So we get negative for cosine of x And then do the do the other the other antiderivatives there we get negative for Cosine of x antiderivative of x to the fourth that's going to be two over five x to the fifth And then we're going to get negative x to the positive one-half power we added one to negative one-half so divide by one-half And so in the end we end up with Negative four cosine of x Plus two fifths x to the fifth Minus well if you divide by one-half that's a two and we get the square root of x That's just the one-half power plus a constant You can leave it as a one-half power if you prefer and this gives us our antiderivative All right, stay tuned for the next video We're going to look at some applications of this antiderivative And see particularly why we care about this plus c so much. See ya