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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 )
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have 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.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Yes (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".

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0After trying it quickly, I don't think the same thing can be applied (not easily, anyway)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Good! :) 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" ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which method do you mean?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/users/watchmath#/users/watchmath/updates/4dbcd050b0ab8b0b5c84818b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0just integration by parts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm, yeah I guess that would be something to try. I'm not going to, though, it looks like it would be horrible!

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0the problems on the exam can be done by parts to after splitting the ln into two ln's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh yes, it would probably work. I'm just not going to do it, because it looks like it would be horribly long

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0hmm in myopinion is not that long :). BTW how did you the problem when on the bottom you have \((x+3)^2\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You want to see the solution (using the substitution method)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Just the rough idea. What do you do with the ln.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But you probably have to them them all in the order they were given (if using the substitution method), because each problem helps the next.

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A 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).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0really you don' need any substitution again? I thought we need to we need to rewrite the ln into \(\ln((x+1)/(x+3))\ln((x+2)/(x+3))\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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)

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0If 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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