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# Integration using long division

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

## Video transcript

see if you can evaluate this integral right over here so I'm assuming you've had a go at it so let's work through this together so you you probably realize that some of the traditional techniques that we've already had in our toolkits don't seem to be directly applicable use substitution and and others and 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 and 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 5 divided by negative 2x plus 2 so let's do a little bit of algebraic long division to actually divide negative 2x plus 2 into X minus 5 to see if we can rewrite this in a way that where we can evaluate the integral so let's do that we're going to take we're going to take X minus 5 so X minus 5 and divide negative 2x plus 2 into that so negative 2x plus 2 so look at the highest degree terms how many times does negative 2x go into X well it's going to go negative 1/2 times negative 1/2 times 2 is negative 1 negative 1/2 times negative 2x is just going to be positive X just like that and now we want to subtract this yellow expression from this blue expression and so let's just let me just take the negative of this and then add so I'm just going to take the negative of it and add and so we are left with negative 5 plus 1 is negative 4 so you could say negative 2x plus 2 goes into X minus 5 negative 1/2 times with negative 4 leftover so we can rewrite this integral our original integral as we can rewrite it as negative 1/2 negative 1/2 minus 4 minus 4 over minus 4 over negative 2 X plus 2 DX DX now let's see it 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 have all these negatives that always unnecessarily complicates things so let's actually divide the numerator and the denominator by negative two so what are we going to have then so if we divide the numerator by negative two of this view 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 and then two divided by negative two is going to be minus one so our original integral once again this is just all algebra everything we've done so far is algebra we've just rewritten it using a little bit of algebraic long division our original integral has simplified to negative one half and some might argue that's not simplified but it's actually much more useful for finding the integral negative 1/2 plus two over X minus 1 DX now how do we evaluate this well the antiderivative of the antiderivative of negative 1/2 is pretty straightforward that's just going to be negative 1/2 X negative 1/2 X but what's the antiderivative of 2 over X minus 1 and you might be able to do this in your head the derivative of X minus 1 is just 1 so you could say that the derivative is sitting there and so we can essentially do u substitution in our heads and say ok let's just take the antiderivative I guess you could say with respect to 2x minus 1 which would be the natural log of the absolute value of X minus 1 if all of that sounds really confusing I'll actually do the u substitution so if I would have if I were to just if I were just trying to evaluate if I were just trying to evaluate the integral 2 over X minus 1 DX I could see okay the derivative of X minus 1 is just 1 so I could say U is equal to X minus 1 and then D U is going to be equal to DX and so this is going to be we can rewrite in terms of U as two I'll just take the constant out two times the N integral of 1 over u D U which we know is two times the natural log of the absolute value of U plus C and in this case we know that U is X minus one so this is equal to two times the natural log of X minus one plus C and so that's what 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 and the plus C doesn't just come from this one this we in general when we're taking the integral of the whole thing there could be some constant because obvious if we go the other way we take the derivative of the constant will go away so let me just put the plus C right over there and we are done
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