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Video transcript

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 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 a over X plus B B over X minus 6 and if I actually had to add these two I would try to find a common denominator and the the best common denominator is just take the product of these two expressions so I could multiply the numerator of the denominator of this first term by X minus 6 X minus 6 and then the numerator of the denominator of the second term I can multiply it 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 6 a plus BX plus BX over over X times X minus 6 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 A's and 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 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 is that the X terms right over here must cancel out since we have no X terms here so we have a X plus VX 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 six a is equal to one or a is equal to divide both sides by negative six negative one-sixth and then if a is negative one-sixth 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 one six 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 antiderivative of this right over here well the antiderivative 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 antiderivative 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 ok I have that derivative laying around and so this is so I can just take the antiderivative 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 how to figure out this you didn't just have to have that insight they're like okay how do I evaluate this and antiderivative well they're telling us did you use partial fraction decomposition
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