How prove this integral $\int_0^1\frac{(\arctan{x})^2\ln({x+1/x+2})}{(1+x)^2}dx$

Question

$$I=\int_0^1\frac{(\arctan{x})^2\ln({x+1/x+2})}{(1+x)^2}dx=-\dfrac{\pi^3}{96}+\dfrac{5\pi}{16}\ln^22-\dfrac{\pi}{4}G-G+\dfrac{\pi}{2}\ln2+\dfrac{7}{16}\zeta(3)$$ Where G is the Catalan's Constant. Using integration by parts we have: $$v=-\dfrac{1}{1+x}, du=[2\dfrac{\arctan{x}\ln(x+1/x+2)}{1+x^2}-\dfrac{(1-x)}{x(1+x)}{(\arctan{x})^2}]dx$$. $$\dfrac{(1-x}{x(1+x)^2}=-\dfrac{1}{1+x}-\dfrac{2}{(1+x)^2}+\dfrac{1}{x}$$ $$\int_0^1\dfrac{(\arctan{x})^2}{x}dx=\dfrac{\pi}{2}G-\dfrac{7}{8}\zeta(3),\int_0^1\frac{(\arctan{x})^2}{(1+x)^2}dx=-\dfrac{G}{2}+\dfrac{\pi}{4}\ln2,\int_0^1\dfrac{(\arctan{x})^2}{1+x}dx=\dfrac{\pi}{4}G-\dfrac{21}{32}\zeta(3)+\dfrac{\pi^2}{32}\ln2$$ $$ And \int_0^1\dfrac{(1-x)}{x(1+x)^2}{(\arctan{x})^2}dx=\dfrac{\pi}{4}G-\dfrac{7}{32}\zeta(3)-\dfrac{\pi^2}{32}\ln2+G-\dfrac{\pi}{2}\ln2$$ How to calculate$$\int_0^1\frac{\arctan{x}\ln({x+1/x+2})}{(1+x)(1+x^2)}dx??$$

Answer 1

Break up the remaining integral as follows

\begin{align} &\int_0^1\frac{\tan^{-1}{x}\ln({x+\frac1x+2})}{(1+x)(1+x^2)}dx\\ =& \ \frac12 \int_0^1\frac{(1-x)\tan^{-1}{x}\ln({x+\frac1x+2})}{(1+x)(1+x^2)}dx+\frac12 \int_0^1\frac{\tan^{-1}{x}\ln({x+\frac1x+2})}{1+x^2}dx\\ =&\ \frac12\left( -\frac{\pi^3}{96}+\frac{5\pi}{16}\ln^22\right) + \frac12\left(\frac7{32}\zeta(3)+\frac{\pi^2}{32}\ln2 \right)\\ = &\ \frac7{64}\zeta(3)- \frac{\pi^3}{192} + \frac{\pi^2}{64}\ln2 + \frac{5\pi}{32}\ln^22 \end{align}