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.

@Pratik Roy
–
Ok, here it is: let
\[I=\int_{1/2014}^{2014} \dfrac{\arctan(x)}{x} \, dx\]
Use the sub u=1/x to get
\[I=\int_{1/2014}^{2014} \dfrac{arccot(x)}{x} \, dx\]
add these to get
\[2I=\int_{1/2014}^{2014} \dfrac{\arctan(x)+arccot(x)}{x} \, dx\]
Now apply integration by parts:
\[\int f'g=fg-\int fg'\]
\[f'=\dfrac{1}{x}\to f=\ln(x), g= \arctan(x)+arccot(x)\to g'=0\]
now the integral is just
\[2I=(\arctan(x)+arccot(x))ln(x)|_{1/2014}^{2014}-\int 0 dx=(\arctan(x)+arccot(x))ln(x)|_{1/2014}^{2014}\]
I think you can put the limits yourself now.

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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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