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Let k equals 6, so that f of x is equal to 1 over x squared minus 6x. Find the partial fraction decomposition for the function f, find the integral of f of x dx.

And so, let's first think 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 6, and 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 6.

And 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 can multiply the numerator and the denominator of this first term by x minus 6, and then the numerator and denominator of the second term I can multiply by the denominator of the other one, so times x and times x.

And so that would give us— let’s see, if I distribute the a, it would give us a x minus 6a plus bx plus b over x times x minus 6. If this looks completely foreign to you, I encourage you to watch the videos on Khan Academy on partial fraction decomposition.

All right, so let's see if we can— so what we want to do is we want to solve for the a's and we want to solve for the b's. And so let’s see if we see that these have to add up to 1 over x times x minus 6. So, these have to add up— I’m just going back to this— these have to add up to the numerator, so it has to add up to 1 over x times x minus 6.

And so, this numerator has to add up to 1. So, what we can see is that the x terms right over here must cancel out, since we have no x terms here. So, we have a x plus b x must be equal to zero, or you could say, well that means that 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 6a is equal to one, or a is equal to— divide both sides by negative 6— negative 1/6. And then if a is negative 1/6, well b is going to be the negative of that. We have negative 1/6 plus b— I’m just substituting a back into that equation— need to be equal to 0. And so, add 1/6 to both sides, you get b is equal to 1/6.

So, I can decompose f of x. I can decompose f of x as being equal to a over x, so that’s negative 1/6 over x. I just write it that way. I could write it as negative 1 over 6x or something like that, but I’ll just write it like this just to be clear that this was our a plus b, which is 1/6 over x minus 6 over x minus 6.

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 fraction decomposition is going to be valuable. That’s going to be the indefinite integral of negative 1/6 over x plus positive 1/6 over x minus 6, and then we have dx.

Well, what’s the anti-derivative of this right over here? Well, the anti-derivative of 1 over x is the natural log of the absolute value of x. And so, we can just say this is going to be negative 1/6 times the natural log of the absolute value of x. That’s the anti-derivative of this part, and then plus 1/6— plus 1/6.

You could do u substitution, but you could just say, hey look, the derivative of this bottom part x minus 6, that’s just 1. And so you could say that, okay, I have that derivative laying around and so this is— so I can just take the anti-derivative with respect to that. And so that’s going to be 1/6 times the natural log of the absolute value of x minus 6.

And then I have— 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 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—they're like, okay, how do I evaluate this anti-derivative? Well, they’re telling us to use partial fraction decomposition.