# Integration

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

Note by Pratik Roy
3 years, 2 months ago

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

- 3 years, 2 months ago

- 3 years, 2 months ago

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.

- 3 years, 2 months ago

thanks

- 3 years, 2 months ago

how can i calculate the integral if the limits are given?

- 3 years, 2 months ago