you 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
$I=\dfrac{i}{2}\int \dfrac{ \ln(1+ix)-\ln(1-ix)}{x}dx= \dfrac{i}{2}(li_2(-ix)-li_2(ix))+C$

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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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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