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## Integrating functions using long division and completing the square

Current time:0:00Total duration:5:18

# Integration using long division

AP Calc: FUN‑6 (EU), FUN‑6.D (LO), FUN‑6.D.3 (EK)

## Video transcript

- [Voiceover] See if you can evaluate this integral right over here. Assuming you had a goal at it so let's work through this together. You probably realized that
some of the traditional techniques that we've already had in our tool kits don't seem to be directly applicable, u-substitution and others. The key here to realize
is we have a rational expression here where the
numerator has the same degree or higher than the denominator. In this case, the numerator and the denominator have the same degree. Whenever you see something
like that, it's probably a good idea to divide the denominator
into the numerator. That's what this rational expression could be interpreted as. X minus five, divided by
negative two x plus two. So it's a little bit of
algebraic long division. To actually divide negative two x plus two into x minus five is to
see if we can rewrite this in a way where we can
evaluate the integral. So let's do that. We're going to take x minus five. X minus five, and divide negative
two x plus two into that. Negative two x plus two. Look at the highest degree terms. How many times does
negative two x go into x? Let's going to go negative 1/2 times. Negative 1/2 times two is negative one. Negative 1/2 times negative two x is this is going to be positive x. Just like that. Now we want to subtract
this yellow expression from this blue expression,
and so let's just, I'll just take the negative
of this and then add. So I'm just going to take
the negative of it and add. We are left with negative five
plus one is negative four. You could say negative two
x plus two goes into x minus five, negative 1/2 times
with negative four left over. So we can rewrite this integral,
our original integral as, we can rewrite it as
negative 1/2 minus four over negative two x plus two, d x. Now let's see, looks like we can simplify this expression a little bit more. The numerator and the denominator, they're both divisible by two. All of these terms are divisible by two. Actually, we all have these negatives that always unnecessarily
complicate things. Let's actually divide the numerator and the denominator by negative two. What are we going to have then? We divide the numerator by negative two. If this is negative four, this is going to become positive two. Then this, if we divide negative two x by negative two, that's
just going to become x. Then two divided by negative
two is going to be minus one. So our original integral... This is just all algebra. Everything we've done so far is algebra. We just rewritten it using a little bit of algebraic long division. Our original integral has
simplified to negative 1/2, and some might argue it's not simplified but it's actually much more
useful for finding the integral. Negative 1/2 plus two
over x minus one, d x. Now, how do we evaluate this? Well, the antiderivative of negative 1/2 is
pretty straight forward. That's just going to be negative 1/2 x, plus the entire derivative
of two over x minus one. You might reduce here ahead. The derivative of x minus one is just one so you could say that the
derivative is sitting there. We can essentially use
substitution in our heads and say, "Okay, let's just
take the entire derivative, "we could say with respect
to x minus one which will be "the natural log of the
absolute value of x minus one." If all of that sounds really confusing, I'll let you do the u-substitution. If I were just trying to evaluate the integral two over x minus one, d x, I could see, okay, the derivative
of x minus one is just one so I could say u is equal to x minus one, and then d u is going to be equal to d x. This is going to be, we can
rewrite in terms of u as two, I'll just take the constant
out two times the integral of one over u, d u, which
we know as two times the natural log of the
absolute value of u plus c. In this case, we know
that u is x minus one. This is equal to two times the natural log of x minus one plus c. That's what we're going
to have right over here. So plus two times the natural log of the absolute value
of x minus one plus c. The plus c doesn't just
come from this one, this general overtaking the
integral of the whole thing. It could be some constant
because if we go the other way, we take the derivative and
the constant will go away. Let me just put the
plus c right over there, and we are done.

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