Let's use use substitution to do this problem. So if we let U equal 2X cubed then DU is going to be 6 X squared DX. So we have a three X squared DX. We could think of 1/2 DU equaling 3X squared DX, just taking each side and multiplying by 1/2. So right here is my three X squared DX. So that's going to get replaced by 1/2. I can take constants in and outside of the integral. So then we've got this square root of U ^2 -, 19. We have an X ^3. Well, if we look at this, we could think of U / 2 as X ^3. So this X ^3 U / 2, which is actually going to let those twos cancel because we'd have 1/2. If I have 1 / 1/2, that's really multiplied by two DU over U square root U ^2 -, 19. So then this is really just a formula. If I have DU over U square root U ^2 -, 19, this here is our a ^2 in our formula. So we're going to have 1 / sqrt 19 secant inverse of 2X cubed. Actually, let's do it in steps U / sqrt 19 + C But we didn't start with U's, we started with XS. So when we take EU out and we put in what it equaled, we'd have the 2X cubed over squared of 19 + C.