bmanmcfly
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I figured I would just add this new problem over here, rather than starting a new thread.
Im looking to solve integration leading to arctan or arcsin results.
[math]\int_{1}^{e}\frac{3dx}{x(1+\ln(x)^2})[/math]
Looking at this, it feels like this has an arctan in the result, but I would have to multiply the x with the [math](1+\ln(x)^2)[/math] 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 [math]tan^{-1}, Cot and arctan[/math] it seems to me that these are just different ways of saying cos/sin. Is there something else significant that I'm ignorant about?
Im looking to solve integration leading to arctan or arcsin results.
[math]\int_{1}^{e}\frac{3dx}{x(1+\ln(x)^2})[/math]
Looking at this, it feels like this has an arctan in the result, but I would have to multiply the x with the [math](1+\ln(x)^2)[/math] 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 [math]tan^{-1}, Cot and arctan[/math] it seems to me that these are just different ways of saying cos/sin. Is there something else significant that I'm ignorant about?