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You are not logged in. #1 20130613 11:57:26
Integration by substitution and natural logarithmsHello again Well, this one wasn't that easy, but my working so far (which I think is on the right track, but maybe here's where my problem is after all) goes like this: From which I get: But the fact that I can't get the correct answer from here and that wolfram alpha tells me that: Rather makes me suspect that I'm wrong about this I'm sure it must have something to do with the fact that the natural logarithm isn't defined for z ≤ 0, but i just can't seem to work out why this is the correct answer and my textbook disperses it's calculus over the course of the book just a bit so it's not that easy to find this information, at least not without starting from page one and working through to the very end Last edited by Au101 (20130613 12:00:07) #2 20130613 12:27:53
Re: Integration by substitution and natural logarithmsHi; for both. Not worrying about the 1 / 2. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #5 20130613 12:56:18
Re: Integration by substitution and natural logarithmsAn antiderivative is a class of functions, there can be more than one. In definite integration it all gets absorbed into the constant of integration. This is how I understand it. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #7 20130613 13:08:58
Re: Integration by substitution and natural logarithmsHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #8 20130613 13:27:21
Re: Integration by substitution and natural logarithmsHmmm...that's what i was trying to do, well, I really need to get off to bed, but for what it's worth, our old friend Wolfram gives, as its answer to This: Which I can get from my answer, so I guess it's just a question of laws of logs, I don't suppose anyone has any ideas? Last edited by Au101 (20130613 13:28:10) #9 20130613 13:31:53
Re: Integration by substitution and natural logarithmsThey are all antiderivatives of that integrand. This can be proven by differentiation. I would do it the way you did. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #10 20130613 17:18:27
Re: Integration by substitution and natural logarithmshi Au101
Strictly, it is the expression within the log that isn't defined (eg. 1z) And log has been extended into complex numbers to allow for the log of a negative. This is just as well in view of what I do below. This may seem strange but (i) you are right to think it's all down to the laws of logs and (ii) logs still obey those laws even for values that are undefined in real numbers. For example even though you might think those negative logs shouldn't 'exist'. And all this means that your first answer is correct. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #12 20130614 00:18:14
Re: Integration by substitution and natural logarithmsI think those two answers are algebraically equal. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #14 20130614 00:27:47
Re: Integration by substitution and natural logarithmsHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #15 20130616 05:38:51
Re: Integration by substitution and natural logarithms
As an aside, you can speed up the partial fractions bit by noting that in other words, whenever the products in the denominator differ by one, you can do this. In general; with x ≠ a,b and a ≠ b. 