A inverse Trigonometric multiple Integrals

Question

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

Answer 1

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