Rewriting before integrating | AP Calculus AB | Khan Academy

Let's say that we wanted to take the indefinite integral of ( x^2 \times (3x - 1) , 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 happens if we distribute this ( x^2 )? 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^2 \times 3x = 3x^3 ) and then negative 1 times ( x^2 = -x^2 ), and then that times ( dx ).

Now, this is pretty straightforward to evaluate. This is going to be equal to the anti-derivative of ( x^3 ), which is ( \frac{x^4}{4} ). So this is going to be ( \frac{3x^4}{4} ). I could write it that way or let me just write it ( \frac{x^4}{4} ).

Then the anti-derivative of ( x^2 ) is ( \frac{x^3}{3} ), so minus ( \frac{x^3}{3} ). 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 easy to evaluate the anti-derivative.

Let's do another example. Let's say that we want to take the indefinite integral of ( \frac{x^3 + 3x^2 - 5}{x^2} , 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^2 )? Well then this thing is going to be equal to, put some parentheses here: ( \frac{x^3}{x^2} = x ), ( \frac{3x^2}{x^2} = 3 ), and then ( \frac{-5}{x^2} ) you could just write that as ( -5x^{-2} ).

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 ( \frac{x^2}{2} ), ( \frac{x^2}{2} + 3x ), and the anti-derivative of ( -5x^{-2} ).

So we would increment the exponent by 1 (positive 1) and then divide by that value. So there would be ( -5x^{-1} ), we're adding one to 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'd have a minus and then you're dividing by negative one, so it's really just going—you can rewrite it like this: ( +5x^{-1} ). You could take the derivative of this to verify that it would indeed give you that, and of course we can't forget our ( +C ). Never forget that if you're taking an indefinite integral.

All right, let's just do one more for good measure. Let's say we're taking the indefinite integral of ( \sqrt[3]{x^5} , dx ). Pause the video and see if you can evaluate this.

Try to write a little bit neater: ( \sqrt[3]{x^5} , 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^{\frac{5}{3}} , dx ). I just rewrote the cube root as the ( \frac{1}{3} ) power ( dx ), which is the same thing as the integral of ( x^{\frac{5}{3}} ).

Many of you might have just gone straight to this step right over here. Then once again, we just have to use the reverse power rule. This is going to be ( x^{\frac{5}{3}} ).

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. So, ( x^{\frac{5}{3}} , dx ). We increment this ( \frac{5}{3} ) by 1 or we can add ( \frac{3}{3} ) to it, so it's ( x^{\frac{8}{3}} ).

Then we divide by ( \frac{8}{3} ) or multiply by its reciprocal. So we could just say ( \frac{3}{8} \times x^{\frac{8}{3}} ).

And of course, we have our ( +C ) and verify this. If you use the power rule here, you'd have ( \frac{8}{3} \times \frac{3}{8} ) would just give you a coefficient of 1, and then you decrement this by 1.

You get to ( \frac{5}{3} ), 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.