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## Integrating with u-substitution

# 𝘶-substitution: special application

## 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.