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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

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

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

After trying it quickly, I don't think the same thing can be applied (not easily, anyway)

4. watchmath

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

Which method do you mean?

6. watchmath
7. anonymous

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

...:(

9. watchmath

just integration by parts

10. anonymous

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

the problems on the exam can be done by parts to after splitting the ln into two ln's

12. anonymous

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

hmm in myopinion is not that long :). BTW how did you the problem when on the bottom you have $$(x+3)^2$$

14. anonymous

You want to see the solution (using the substitution method)?

15. anonymous

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

Just the rough idea. What do you do with the ln.

17. anonymous

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

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

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

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

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

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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