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Topic review (newest first)

2013-06-16 05:38:51

Au101 wrote:

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.

2013-06-14 00:27:47


I filled in post #12 with the proof.

2013-06-14 00:26:29

i think you're right, thank you very much bobbym smile

2013-06-14 00:18:14

I think those two answers are algebraically equal.

2013-06-13 22:44:21

Thank you very much to both bobs, tehe, I only wonder if anybody knows how:

Could have been obtained. If only out of interest

bob bundy
2013-06-13 17:18:27

hi Au101

you wrote:

the natural logarithm isn't defined for z ≤ 0

Strictly, it is the expression within the log that isn't defined (eg. 1-z)  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.  smile

Your expression and Wolfram's are the same except for a constant:

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.  dizzy

For example

even though you might think those negative logs shouldn't 'exist'.  smile

And all this means that your first answer is correct.  smile smile


2013-06-13 13:31:53

They are all antiderivatives of that integrand. This can be proven by differentiation. I would do it the way you did.

2013-06-13 13:27:21

Hmmm...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


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?

2013-06-13 13:08:58


When I differentiate that book answer it is correct. I check integrations by differentiating, use Wolfram if you need to. You could also use it to check every step of your solution.

2013-06-13 13:04:29

Okay, thanks a lot bobbym, i'll give it a try in the morning, perhaps i just made a mistake when i put:

Back into the equation, which is why I couldn't get the answer which the book has:

2013-06-13 12:56:18

An 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.

Use your answer and working. Mathematica uses different algorithms than we use by hand. There will sometimes be differences.

2013-06-13 12:52:32

Sorry bobbym, yes, i agree, i get the same for both.

2013-06-13 12:51:30

I hadn't, but having done so, my answer:

Gives me:

And wolfram's gives:

Which, surely, is equivalent to:

2013-06-13 12:27:53


Did you differentiate your answer and Wolframs?

I got

for both. Not worrying about the 1 / 2.

2013-06-13 11:57:26

Hello again smile

I've been doing a bit of practice of integration by substitution and it was all going very well until I got to a question involving the natural logarithm and I got a bit stuck because it's been rather a long time...Anyway, i'm complaining again tongue Here's the question:

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 tongue 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 tongue

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