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watchmath

  • 5 years ago

For INewton and everybody interested in it Compute \[\int_0^1\frac{\ln((x+1)/(x+2))}{(x+3)^3}\,dx\] (yes, it is \((x+3)^3\) now )

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  1. anonymous
    • 5 years ago
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    Have you solved it? I assume the same substitution reduces it to something manageable, but another substitution is necessary? I'll probably have a look in a minute.

  2. watchmath
    • 5 years ago
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    Yes (some rough sketch :) ). I don't know the numerical value of it though. I am just wondering, how one solve this if you know that "method".

  3. anonymous
    • 5 years ago
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    After trying it quickly, I don't think the same thing can be applied (not easily, anyway)

  4. watchmath
    • 5 years ago
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    Good! :) Here is the "method" that I was talking about (from INewton old post). But do you see that the problems on the exam actually can be solved without that "method" ?

  5. anonymous
    • 5 years ago
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    Which method do you mean?

  6. watchmath
    • 5 years ago
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    http://openstudy.com/users/watchmath#/users/watchmath/updates/4dbcd050b0ab8b0b5c84818b

  7. anonymous
    • 5 years ago
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    I can't see it right now (but a substitution seems the obvious thing to do anyway, even if the question is a bit convoluted). What method do you have in mind to do it otherwise?

  8. anonymous
    • 5 years ago
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    ...:(

  9. watchmath
    • 5 years ago
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    just integration by parts

  10. anonymous
    • 5 years ago
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    Hmm, yeah I guess that would be something to try. I'm not going to, though, it looks like it would be horrible!

  11. watchmath
    • 5 years ago
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    the problems on the exam can be done by parts to after splitting the ln into two ln's

  12. anonymous
    • 5 years ago
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    Oh yes, it would probably work. I'm just not going to do it, because it looks like it would be horribly long

  13. watchmath
    • 5 years ago
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    hmm in myopinion is not that long :). BTW how did you the problem when on the bottom you have \((x+3)^2\)

  14. anonymous
    • 5 years ago
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    You want to see the solution (using the substitution method)?

  15. anonymous
    • 5 years ago
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    You can factorise the top of the log and split it up into two. One of them is the same as the first part, and the other one falls to a similar substitution.

  16. watchmath
    • 5 years ago
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    Just the rough idea. What do you do with the ln.

  17. anonymous
    • 5 years ago
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    But you probably have to them them all in the order they were given (if using the substitution method), because each problem helps the next.

  18. watchmath
    • 5 years ago
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    I understand each step. I just want to see how other people see it. I want to see how to deal with the last problem since there is no \(x+3\) inside the ln this time.

  19. anonymous
    • 5 years ago
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    A hint is to consider the second part (with the (x+3)^2 in the ln) - both the terms once you factorise/split up - and how you can use them to find the last one (you don't need to do any more substitution).

  20. watchmath
    • 5 years ago
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    really you don' need any substitution again? I thought we need to we need to rewrite the ln into \(\ln((x+1)/(x+3))-\ln((x+2)/(x+3))\)

  21. anonymous
    • 5 years ago
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    Yeah, that's exactly what you need to do! (I meant you don't need to substitute again because you have already worked out both of them)

  22. watchmath
    • 5 years ago
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    If I don't know the (rational) substitution method We can compute \(\int \frac{1}{(x+3)^2}\ln(x+1)\,dx=-(x+3)^{-1}\ln(x+1)+\int \frac{1}{(x+1)(x+3)}\, dx\) which is not bad at all. Then we can do the same for \(\int \frac{1}{(x+3)^2}\ln(x+2)\,dx\)

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