\[\large \int \dfrac{\arctan(x)}{x} \, dx\] Can anyone tell how this integration can be done?

\[\large \int \dfrac{\arctan(x)}{x} \, dx\] Can anyone tell how this integration can be done?

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TopNewestyou need knowledge of euler's formula and polylogarithm. first we write \[\tan(y)=x\to x=\arctan(x)=y\] or \[x=\dfrac{\sin(y)}{\cos(y)}=-i\dfrac{e^{2iy}-1}{e^{2iy}+1}\] So \[y=\dfrac{i}{2} \ln(1+ix)-\dfrac{i}{2} \ln(1-ix)\] so, \[\arctan(x)=\dfrac{i}{2} \ln(1+ix)-\dfrac{i}{2} \ln(1-ix)\] putting this in the integral \[I=\dfrac{i}{2}\int \dfrac{ \ln(1+ix)-\ln(1-ix)}{x}dx= \dfrac{i}{2}(li_2(-ix)-li_2(ix))+C\]

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can you please suggest me from where i could learn more about doing these type of integrations.

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from here try the integration tricks wikis and advance integration. while this is not complete yet the contents there are good enough for this problem.

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Comment deleted Apr 06, 2016

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PS dont use all caps it is considered rude.

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Comment deleted Apr 06, 2016

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\( tan^{-1}(x) + cot^{-1}(x) = \dfrac{\pi}{2} \)

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