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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 11

Lesson 2: AP Calculus BC 2015- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c

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# 2015 AP Calculus BC 5d

Partial fraction expansion.

## Want to join the conversation?

- Couldn't you use the natural log properties to simplify it to be 1/6ln abs(x-6/x) + c?(5 votes)
- Yes, you can use the properties of logarithms.(4 votes)

## Video transcript

- [Voiceover] Let K equal six,
so that F of X is equal to one over X-squared minus six X. Find the partial fraction
decomposition for the function F. Find the integral of F of X D-X. And so let's first thinl about the partial fraction
decomposition for the function F. So, F of X, I could rewrite it where I factor the denominator. If I factor out an X, I
get X times X minus six. So, I can rewrite this as... And this is where I'm going to decompose it into partial fractions. A over X plus B over X minus six. If I actually had to add these two, I would try to find a common denominator. The best common denominator is just to take the product
of these two expressions. So I could multiply the
numerator and denomenator of this first time by X minus six. X minus six. And the numerator and
denominator of the second term, I can multiply it by the
denominator of the other one. So times X and times X. So that would give us, let's see if I distribute the A, it would give us A X minus six A plus B X over X times X minus six. If this looks completely foreign to you, I encourage you to watch
the videos on Kahn Academy on partial fraction decomposition. So let's see if we can, so what we wanna do is solve
for the A's and the B's. And so let's see, if we see that these have to add up to one over X times X minus six. So these have to add up. I'm just going back to
this, these have to add up. The numerator. So it has to add up to one over X times X minus six. So this numerator has to add up to one. So what we can see is is that the X terms right over here must cancel out since
we have no X terms here. We have A X plus B X
must be equal to zero. Or you could say, well that means A plus B is equal to zero. So we took care of that
term and that term. And then we know, that this
must be the constant term that adds up to one or
that is equal to one. And so we also know that
negative six A is equal to one. Or A is equal to, divide
both sides by negative six, negative one sixth. And if A is negative one sixth, well B is going to be
the negative of that. And negative one sixth plus B. I'm just substituting A
back into that equation. Need to be equal to zero. And so add one sixth to both sides, you get B is equal to one sixth. So I can decompose F of X. I can decompose F of X as
being equal to A over X So that's negative one sixth over X. I just write it that way. I could write it as
negative one over six X or something like that. But I'll just write it
like this, just be clear that this was our A. Plus B, which is one
sixth over X minus six. So that right over there, that's the partial fraction decomposition for our function F. And if I want to evaluate the integral. So the integral of F of X,
the indefinite integral. Well, that's where this
partial fractions decomposition is going to be valuable. That's going to be the indefinite integral of negative one sixth over
X plus positive one sixth over X minus six, and then we have D X. Well what's the anti-derivative
of this right over here? Well the anti-derivative of one over X is the natural log of
the absolute value of X. And so we can just this is going to be negative one sixth times
the natural log of the absolute value of X. Anti-derivative of this part. Then plus one sixth. You could do u-substitution
but you could just say "Hey, look the derivative of
this bottom part, X minus six." That's just one. So you could say that I have
that derivative laying around. So I can just take the
anti-derivative with respect to that. So that's going to be one
sixth times the natural log of the absolute value of X minus six, and then I have plus C. Don't forget this is an
indefinite integral over here. And then you're done. And you see that partial
fraction decomposition was actually quite useful. So they were helping us how to figure out this. You didn't just have to have that insight that, okay how do I evaluate
this anti-derivative. Well, the they're telling us to use partial fraction decomposition.