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Im looking to solve integration leading to arctan or arcsin results.

\(\displaystyle \int_{1}^{e}\frac{3dx}{x(1+\ln(x)^2})\)

Looking at this, it feels like this has an arctan in the result, but I would have to multiply the x with the \(\displaystyle (1+\ln(x)^2)\) and then would figure out a difference of squares to figure out the integral...

that ln(x) is what is screwing me up...

So, I guess I'm asking how working with the ln(x)^2 would be different from working with just an x.

Side question, what is the difference really between \(\displaystyle tan^{-1}, Cot and arctan\) it seems to me that these are just different ways of saying cos/sin. Is there something else significant that I'm ignorant about?