Let $n\geq1$ be an integer and let
$$I_n=\int\limits_{0}^{\infty}\dfrac{\arctan x}{(1+x^2)^n} \,\mathrm dx$$
Prove that
$$\sum\limits_{n=1}^{\infty}\dfrac{I_n}{n}=\dfrac{\pi^2}{6} \tag{1}$$
$$\int\limits_{0}^{\infty} \arctan x\cdot\ln\left(1+\frac{1}{x^2}\right) \,\mathrm d x=\dfrac{\pi^2}{6} \tag{2}$$
I have an idea that dominated convergence theorem may help here. But I am not getting the proper way.
You can use the monotone convergence theorem to show that the series and the integral are equal: \begin{align} \sum \limits_{n=1}^\infty \frac{I_n}{n} &\stackrel{\text{MCT}}{=} \int \limits_0^\infty \arctan(x) \sum \limits_{n=1}^\infty \frac{1}{n (1+x^2)^n} \, \mathrm{d} x = \int \limits_0^\infty \arctan(x) \left[-\ln\left(1 - \frac{1}{1+x^2}\right)\right] \, \mathrm{d} x \\ &\hspace{5pt}= \int \limits_0^\infty \arctan(x) \ln \left(1 + \frac{1}{x^2}\right) \, \mathrm{d} x \, . \end{align} The integral can be evaluated by integrating by parts: \begin{align} \int \limits_0^\infty \arctan(x) \ln \left(1 + \frac{1}{x^2}\right) \, \mathrm{d} x &= \int \limits_0^\infty \left[\frac{2 \arctan{x}}{1+x^2} - \frac{\ln(1+x^2)}{x (1+x^2)}\right] \, \mathrm{d} x \\ &\hspace{-11pt}\stackrel{x = \sqrt{\mathrm{e}^t - 1}}{=} \frac{\pi^2}{4} - \frac{1}{2} \int \limits_0^\infty \frac{t}{\mathrm{e}^t - 1} \, \mathrm{d} t \\ &= \frac{\pi^2}{4} - \frac{1}{2} \zeta(2) = \frac{\pi^2}{4} - \frac{\pi^2}{12} = \frac{\pi^2}{6} \, , \end{align} The integral over $t$ is found using the geometric series and, once again, the monotone convergence theorem. I have not managed to evaluate the sum directly, however.