Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
You are not logged in.
Post a reply
Topic review (newest first)
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.
i think you're right, thank you very much bobbym
I think those two answers are algebraically equal.
Thank you very much to both bobs, tehe, I only wonder if anybody knows how:
Could have been obtained. If only out of interest
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.
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.
even though you might think those negative logs shouldn't 'exist'.
And all this means that your first answer is correct.
They are all antiderivatives of that integrand. This can be proven by differentiation. I would do it the way you did.
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?
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:
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.
Sorry bobbym, yes, i agree, i get the same for both.
I hadn't, but having done so, my answer:
And wolfram's gives:
Which, surely, is equivalent to:
for both. Not worrying about the 1 / 2.
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