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# 𝘶-substitution: special application

Using 𝘶-substitution in a situation that is a bit different than "classic" 𝘶-substitution. In this case, the substitution helps us take a hairy expression and make it easier to expand and integrate. Created by Sal Khan.

## Video transcript

We're faced with the indefinite integral of x plus 3, times x minus 1 to the fifth dx. Now, we could solve this by literally multiplying out what x minus 1 to the fifth is. Maybe using the binomial theorem, that would take a while. And then we'd multiply that times x plus 3, and we'd end up with some polynomial. And we could take the antiderivative that way. Or we could maybe make a substitution here that could simplify this expression here. Make it something that's a little bit easier to take the antiderivative of. And this isn't going to be kind of the more traditional u substitution, where we just set u equal to something and see if its derivative is there. But it's a kind of a form of u substitution, where we do set u equal to something and see if it simplifies our expression in a way that, I guess, simplifies it. So let's try things out. So we have this x minus 1 to the fifth. That would be a pain to expand out. It would be nice if this was just a u to the fifth. So let's just set this, let's just set this to be equal to u. So let's set u as equal to x minus 1. And in that case, du is equal to dx. We could write du/dx is equal to 1, derivative of x, derivative of negative 1 is just 0. And these two are completely equivalent statements. And so if we did that, how could we rewrite this entire expression? Well, it would be equal to the integral of-- well, we have x plus 3 right over here, this is neither just u nor is it du. So let's think about how, what we can do here. Well, we could say, if u is equal to x minus 1, we could add 1 to both sides of this equation. And we could say u plus 1 is equal to x. And so for x, we can substitute that with u plus 1. So let's do that. So we're kind of back substituting in for x. So x is equal to u plus 1. And I'm just trying to see if I can do something to simplify this expression. So x is u plus 1, then we have our plus 3 there, times x minus 1 to the fifth. x minus 1 was u, that's the simplification we wanted to make. So times u to the fifth power. And dx is the same thing as du. So du. Now did this get us anywhere? Can we simplify this to a form that it's easy to take the antiderivative of? Well, it looks like it did. Let's see. We can rewrite this as this expression right over here is just u plus 4. Times u to the fifth-- I'll do it all in one color, now-- du. And the reason why this simplified things-- the way this simplified things is taking x minus one to the fifth, that'd be a really hard thing to expand. But u to the fifth is just u to the fifth. And then it just changed this x plus 3 into a u plus 4, which is still a pretty straightforward expression. And now we can just distribute the u to the fifth. So we are left with u to the sixth power plus 4, u to the fifth du. And this is a pretty straightforward thing to take the antiderivative of. Now you might be saying, hey Sal, this was-- how did you know to set u equal to be that. And oftentimes with integration, it's going to take a little bit of trial and error. There's a certain bit of an art to it. But here the realization was well, x minus 1 to the fifth can be really complicated. Maybe u to the fifth might make it a little bit simpler. And that did just happen to work. You could have tried u is equal to x plus 3. But it wouldn't have simplified it as nicely as u equal to x minus 1 did. But let's finish with this integral right over here. So this is going to be equal to the antiderivative of u to the sixth. Well, that's just u to the seventh over seven. Plus the antiderivative of u to the fifth, that's u to the sixth over six. But we have the 4 out here, so it's 4 times u to the sixth over six. And then we have a plus C. And this, the 4u 6 is the same thing as 2/3, so we can rewrite this whole thing as equal to u to the seventh over seven, plus 2/3 u to the sixth, plus C. And now we just have to undo our u substitution. u is equal to x minus 1. So this is going to be equal to x minus 1 to the seventh over 7 plus 2/3 times x minus 1 to the sixth, plus C. And we're all done. We were able to take a fairly hairy-- or what could have been a hairy thing if we had to expand this out-- and we were able to take the antiderivative by doing a little bit of this u substitution and u back substitution.