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How to evaluate :$$\int\cot^{-1} (x^2+x+1)dx $$

dia
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  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Feb 22 '13 at 15:58

2 Answers2

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Hint: Use integration by parts with v=1, but before doing that note, $\cot^{-1}(x^2+x+1)=\tan^{-1}(\frac1{x^2+x+1})=\tan^{-1}(x+1)-\tan^{-1}(x)$ which simplifies it a lot.

Ishan Banerjee
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Hint : Use integration by parts with $ u=\cot^{-1} (x^2+x+1) $, then you can use partial fraction techniques. Here is the final answer

$$ x\cot^{-1} \left( {x}^{2}+x+1 \right) +\frac{1}{2}\,\ln\left( 1+{x}^{2} \right) -\frac{1}{2}\,\ln \left( {x}^{2}+2\,x+2 \right) +\arctan \left( 1+x \right) $$

Mhenni Benghorbal
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    you must use the same version of Wolfram Alpha as I do! – Ron Gordon Feb 22 '13 at 15:48
  • @rlgordonma: What are you saying? – Mhenni Benghorbal Feb 22 '13 at 15:52
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    I'm saying that I put the integral into WA and got that exact, same result, even precisely as you have it laid out. But perhaps you did do it out and would like to share that info with the rest of us. I, for one, thought the partial fraction decomposition looked very difficult and unwieldy. Ishan's observation, on the other hand, looks like a better way to go. – Ron Gordon Feb 22 '13 at 15:55
  • @rlgordonma: These are different approaches to the problem. The OP can choose any answer he likes, but I am not obliged to use something that you use. I hope this is clear. – Mhenni Benghorbal Feb 22 '13 at 15:58
  • of course! The OP him/herself did a poor job of describing the problem. But you and I worked out the problem the exact same way - we had the same idea (I think we think very much alike, observing your postings). The difference is that I was bogged down in doing this decomposition - it is possible, but extremely tough to do by hand. If you did it out this way as you suggested, and perhaps you did, then you would know that. BTW if you did do it out, then I for one would love to see how you did the decomposition, because I got stuck. – Ron Gordon Feb 22 '13 at 16:02
  • @rlgordonma: I expect the OP to be a student who is learing integration techniques. So the best way to learn this is they have to do some detailed work in order to learn integration techniques and one of these integration techniques is the partial fractions techniques. – Mhenni Benghorbal Feb 22 '13 at 16:07
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    I strongly disagree. By giving such a student a near-impossible example of decomposition, you are discouraging that student from wanting to learn more. I understand that the integral in this question was not that of a first-year calc student (but maybe I am mistaken); still, and I do not mean to brag about my abilities, but if I am getting bogged down working on the problem this way, how is a less experienced student faring? Again, you may have worked this out; if you have, then I for one would love to see it, as presumably would the student. – Ron Gordon Feb 22 '13 at 16:36