𝘶-substitution: multiplying by a constant

Let's take the indefinite integral of the square root of 7x plus 9 dx. So my first question to you is, is this going to be a good case for u-substitution? Well, when you look here, maybe the natural thing to set to be equal to u is 7x plus 9. But do I see its derivative anywhere over here? Well, let's see. If we set u to be equal to 7x plus 9, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to 7. Derivative of 7x is 7. Derivative of 9 is 0. So do we see a 7 lying around anywhere over here? Well, we don't. But what could we do in order to have a 7 lying around, but not change the value of the integral? Well, the neat thing-- and we've seen this multiple times-- is when you're evaluating integrals, scalars can go in and outside of the integral very easily. Just to remind ourselves, if I have the integral of let's say some scalar a times f of x dx, this is the same thing as a times the integral of f of x dx. The integral of the scalar times a function is equal to the scalar times the integral of the functions. So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a 7 showing up? Well, we can multiply and divide by 7. So imagine doing this. Let's rewrite our original integral. So let me draw a little arrow here just to go around that aside. We could rewrite our original integral as being 9 to the integral of times 1/7 times 7 times the square root of 7x plus 9 dx. And if we want to, we could take the 1/7 outside of the integral. We don't have to, but we can rewrite this as 1/7 times the integral of 7, times the square root of 7x plus 9 dx. So now if we set u equal to 7x plus 9, do we have its derivative laying around? Well, sure. The 7 is right over here. We know that du-- if we want to write it in differential form-- du is equal to 7 times dx. So du is equal to 7 times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the 7x plus 9. That is are u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1/7 times the integral of-- and I'll just take the 7 and put it in the back. So we could just write the square root of u du, 7 times dx is du. And we can rewrite this if we want as u to the 1/2 power. It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1/7 times the integral of u to the 1/2 power du. And let me just make it clear. This u I could have written in white if I want it the same color. And this du is the same du right over here. So what is the antiderivative of u to the 1/2 power? Well, we increment u's power by 1. So this is going to be equal to-- let me not forget this 1/7 out front. So it's going to be 1/7 times-- if we increment the power here, it's going to be u to the 3/2, 1/2 plus 1 is 1 and 1/2 or 3/2. So it's going to be u to the 3/2. And then we're going to multiply this new thing times the reciprocal of 3/2, which is 2/3. And I encourage you to verify the derivative of 2/3 u to the 3/2 is indeed u to the 1/2. And so we have that. And since we're multiplying 1/7 times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1/7. So it would get 1/7 times 2/3 is 2/21 u to the 3/2. And 1/7 times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c1 and then I could call this c2, but it's really just some arbitrary constant. And we're done. Oh, actually, no we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2/21 times u to the 3/2. And we already know what u is equal to. u is equal to 7x plus 9. Let me put a new color here just to ease the monotony. So it's going to be 2/21 times 7x plus 9 to the 3/2 power plus c. And we are done. We were able to take a kind of hairy looking integral and realize that even though it wasn't completely obvious at first, that u-substitution is applicable.