Recent content by TomasRiker

Thank you! It is clear to me now.

I understand this - maybe I wasn't clear enough. What surprises me here is that, using those substitutions, we effectively substituted for the entire original integrand. I don't remember having seen such a thing before, and I am wondering whether this is common and in which cases it works.

Hello, I found something surprising (at least to me) while looking at the following integral: \int \sqrt{\frac{e^x-1}{e^x+1}} dx Wolfram Alpha suggests the following substitution as the first step: u = \frac{1}{e^x+1} Which leads to the following integral: \int \frac{\sqrt{1-2u}}{(u-1)u} du The...

Hello, I have created two online calculators which (I hope :smile:) can prove useful: Derivative Calculator (with steps) It can do first, second, third derivative (symbolic) with respect to any variable and optionally simplify input and output. The input formula is being displayed as a...