If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 6

Lesson 8: Finding antiderivatives and indefinite integrals: basic rules and notation: reverse power rule

# Rewriting before integrating

Some indefinite integrals are much simpler to integrate by algebraically rewriting the integrand first.

## Want to join the conversation?

• At , I attempted to rewrite it a different way, which ended up being wrong. Why is it wrong?
I went::
(x^3 + 3x^2 - 5) / x^2
Then I converted the denominator to x^-2, and multiplied it in front of the numerator. Isn't it the same as dividing it?
This resulted: x^-2 (x^3 + 3x^2 - 5)
I multiplied everything inside the brackets with the x to the negative to power and got:
x^-1 + 3x^0 - 5

So integrating it became:
x^0+ 3x - 5x + c

Yikes....
• It's wrong because you didn't multiply everything inside the brackets correctly.
x¯² ( x³ + 3x² - 5) = x + 3 - 5x¯²
Integrating would give x²/2 + 3x + 5/x + C
• At Sal rewrites the equation by dividing every term solely for x^2. However, this new equation would have a new domain that does include x=0. If you were to use defined integral for integration over some range including 0, wouldn't this integral be incorrect?
• The domain of the new equation doesn't include 0, since 0^(-2) is undefined. But even if it did, this wouldn't affect any definite integral we take, since the area above the point 0 contributes no area (it's a rectangle with width 0).
• I'm aware that the derivative of an expression + c is equal to just the expression, but why/how doesn't that change the actual area under the curve?
• Changing the value of +C does change the area under the curve. What it doesn't change is the derivative; shifting the function up and down doesn't affect the slope of the function at a given x-coordinate.
• ok just curious. On the first problem I got ((3x^4)/3)-((x^3)/3) + 2C. I got the 2C from adding F(3x^3) and F(-x^2), each of course coming with their own C.

I get that the C is a random constant so 2C and C are practically the same thing, but does it ever matter?
• For the first example, is there such a thing as a "product rule" for integration?
(1 vote)
• What would you do if you had two functions being divided or multiplied and there was no easy way to simplify the function algebraically? What would you do next, some type of reverse rule?
• can you multiply multiple integrals together? like integral times an integral?
(1 vote)
• If you mean integrating two functions and multiplying them, sure, you could do that. If you mean multiplying the integral symbol with itself, no, that's not a defined operation.
• At Sal mentions the reverse power rule. Can someone explain it?
(1 vote)
• How can you take an antiderivative when the exponent is -1? (like x^-1)
(1 vote)
• Good question! Note that we can’t use the reversed power rule because we would end up dividing by zero.

Instead, we note that the derivative of ln|x| is 1/x, which is x^(-1).

Therefore, since the derivative of a constant term is zero, the most general antiderivative of x^(-1) is ln|x| + c.

Have a blessed, wonderful day!
• at , ∫3 dx= ∫(3*x^0) dx= 3∫(x^0) dx= 3x, is it the correct way to do it and do we have a simpler way to do it?
(1 vote)
• You just perform reverse power rule. The question you have asked is quite simple so you can just answer it in one go instead having various intermediate steps.

e.g. integral 3 dx = 3x + C

## Video transcript

- [Instructor] Let's say that we wanted to take the indefinite integral of X squared times three X minus one DX. Pause this video and see if you can evaluate this. So, you might be saying oh, what kind of fancy technique could I use? But you will see sometimes the fanciest or maybe the least fancy but the best technique is to just simplify this algebraically. So, in this situation what would happens if we distribute this X squared? Well, then we're going to get a polynomial here within the integral, so this is going to be equal to the integral of X squared times three X is three X to the third and then negative one times X squared is minus X squared and then that times DX and now this is pretty straightforward to evaluate. This is going to be equal to the anti-derivative of X to the third is X to the fourth over four so this is going to be three times X to the fourth over four, I could write it that way or let me just write it X to the fourth over four and then the anti-derivative of X squared is X to the third over three, so minus X to the third over three and this is an indefinite integral, there might be a constant there, so let me write that down and we're done. The big takeaway is you just have to do a little bit of distribution to get a form where it's easier to evaluate the anti-derivative. Let's do another example. Let's say that we want to take the indefinite integral of, this is gonna be a hairy expression, so X to the third plus three X squared minus five, all of that over X squared DX. What would this be? Pause the video again and see if you can figure it out. So, once again your brain might want to try to do some fancy tricks or whatever else but the main insight here is to realize that you could just simplify it algebraically. What happens if you just divide each of these terms by X squared? Well, then this thing is going to be equal to, that's in parentheses here, X to the third divided by X squared is just going to be X. Three X squared divided by X squared is just three and then negative five divided by X squared, you could just write that as negative five times X to the negative two power. And so, once again we just need to use the reverse power rule here to take the anti-derivative. This is going to be, let's see the anti-derivative of X is X squared over two, X squared over two plus the anti-derivative of three would just be three X. The anti-derivative of negative five X to the negative two. So, we would increment the exponent by one, positive one and then divide by that value. So, it would be negative five X to the negative one, we're adding one to the negative one, all of that divided by negative one which is the same, we could write it like that. Well, these two would just, you would have minus and then you're dividing by negative one, so it's really just going, we can rewrite it like this plus five X to the negative one and you could take the derivative of this to verify that it would indeed give you that and of course we can't forget our plus C, never forget that if you're taking an indefinite integral. Alright, let's just do one more for good measure. Let's say we're taking the indefinite integral of the cube root of X to the fifth DX. Pause the video and see if you can evaluate this. Try to write it a little bit neater. X to the fifth DX. Pause the video and try to figure it out. So, here the realization is well, if you just rewrite all this as one exponent, so this is equal to the indefinite integral of X to the fifth to the one third, I just rewrote the cube root as the one third power DX, which is the same thing as the integral of X to the, if I raise something to a power and then raise that to a power I can multiply those two exponents, that's just exponent properties, X to the five thirds DX, many of you might have just gone straight to this step right over here and then once again, we just have to use the reverse power rule. This is going to be X to the, we increment this five thirds by one or we can add three thirds to it, so it's X to the eight thirds and then we divide by eight thirds or multiply by its reciprocal so we could just say three eighths times X to the eight thirds and of course, we have our plus C and verify this. If you use the power rule here you'd have eight thirds times the three eights, will just give you a coefficient of one and then you decrement this by three thirds or one, you get to five thirds which is exactly what we originally had. So, the big takeaway of this video, many times the most powerful integration technique is literally just algebraic simplification first.