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How to calculate the closed form of the integral $$\int\limits_0^1 {\frac{{\int\limits_0^x {{{\left( {\arctan t} \right)}^2}dt} }}{{x\left( {1 + {x^2}} \right)}}} dx$$


marked as duplicate by Mark Fantini, Willie Wong, Sujaan Kunalan, graydad, Lord_Farin Mar 31 '15 at 15:38

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  • 1
    $\begingroup$ What is the motivation? Do you know if it has a closed form or just wondering? Personally, I would rather not waste time on such a problem if it is just taken out of the hat (implying that an analytic solution is not likely to exist) so it would be nice if you could clarify. $\endgroup$ – Winther Oct 21 '14 at 23:54

Use integration by parts, taking $\displaystyle \int_{0}^{x} (\arctan t)^2 dt$ as the first function and $ \dfrac{1}{x(x^2+1)} $ as the second function to get:

$ \displaystyle \int \dfrac{dx}{x(x^2+1)} = \dfrac{1}{2} \ln\bigg(\dfrac{x^2}{x^2+1}\bigg)$


$I = \bigg|\dfrac{1}{2} \ln \bigg(\dfrac{x^2}{x^2+1}\bigg) \displaystyle \int_{0}^{x} (\arctan t)^2 dt \bigg|_{0}^{1} - \displaystyle \int_{0}^{1} \dfrac{1}{2} \ln \bigg(\dfrac{x^2}{x^2+1}\bigg) (\arctan x)^2 dx $

= $ \displaystyle \dfrac{1}{2}\int_{0}^{1} (\arctan x)^2 \ln\bigg(\dfrac{1+x^2}{2x^2}\bigg) dx $

I am not able to solve it further

  • $\begingroup$ Mathematica is unable to give any closed form for it $\endgroup$ – Jatin Yadav Oct 23 '14 at 16:25

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