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# Integration using completing the square and the derivative of arctan(x)

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

## Video transcript

all right let's see if we can find the indefinite integral of 1 over 5x squared minus 30x plus 65 DX pause this video and see if you can figure it out alright so this is going to be an interesting one it'll be a little bit hairy but we're gonna work through it together so immediately you might try multiple integration techniques and be hitting some walls and what we're going to do here is actually try to complete the square in this denominator right over here and then by completing the square we're gonna get in the form that it looks like the derivative of arctan and if that's a big hint to you once again pause the video and try to move forward all right now let's do this together so I'm just going to try to simplify this denominator so that my coefficient on my x squared term is a 1 and so I can just factor a 5 out of the denominator and if I did that then this integral will become 1/5 times the integral of 1 over so I factored a 5 out of the denominator so it is x squared minus 6x plus 13 DX and then as I mentioned I'm going to complete the square down here so let me rewrite it so this is equal to 1/5 times the integral from of 1 over and so x squared minus 6x it's clearly not a perfect square the way it's written let me write this plus 13 out here now what could I add and then I'm gonna have to subtract if I don't want to change the value of the denominator in order to make in order to make this part right over here a perfect square well we've done this before you take half of your coefficient here which is negative 3 and you square that so you want to add a 9 here but if you add a 9 then you have to subtract a 9 as well and so this part is going to be X minus 3 squared and then this part right over here is going to be equal to a positive 4 and we of course don't want to forget our DX out here and so let me write it in this form so this is going to be equal to 1/5 time's the integral of one over get myself some space X minus three squared plus four which could also write as plus two squared actually let me do it that way plus two squared DX now many of y'all might already be say hey this looks a lot like arctangent but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved now I'm actually going to do some u substitution in order to do it so the first thing I'm going to do is let's factor a or out of the denominator here so if we do that then this is going to become one fifth times one fourth which is going to be one xx times the integral of one over X minus three squared over two squared and then this is going to be a plus one and of course we have our DX and then we could write this as and I'm trying to just do every step here a lot of these you might have been able to do in your head one over and I'll just write this as X minus three over two squared plus one plus one and then DX and now the u-substitution is pretty clear I am just going to make the substitution that U is equal to X minus three over two or we could even say that's U is equal to one half X minus three halves that's just X minus 3 over 2 and D U is going to be equal to 1/2 DX and so what I can do here is actually let me start to re-engineer this integral a little bit so that we see a 1/2 here so if I make this a 1/2 and then I multiply the outside by 2 so I divided by 2 multiplied by 2 is one way to think about it this becomes 1/10 and so doing my u substitution I get 1/10 that's that one tenth there times the integral of well I have 1/2 X right over here which is the same thing as D u so I could put the D U either in the numerator I could put it out here and then I have one over this is U squared u squared plus one now you might immediately recognize what's the derivative of arctan of you well that would be 1 over u squared plus one so this is going to be equal to one tenth times the arc tangent of U and of course we can't forget our big constant C because taking an indefinite integral and now we just want to do the reverse substitution we know that U is equal to this business right over here so we deserve a little bit of a drumroll this is going to be equal to one tenth times the arc tangent of U well U is just X minus three over two which could also be written like this so arc tan of X minus three over two and then plus C and we are done
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