<s>[INST] A inverse Trigonometric multiple Integrals How to calculate the closed form of the integral
$$\int\limits_0^1 {\frac{{\int\limits_0^x {{{\left( {\arctan t} \right)}^2}dt} }}{{x\left( {1 + {x^2}} \right)}}} dx$$
 [/INST] 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)$
Hence,
$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
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