$$\begin {aligned} \int _{0}^{\infty }\frac {\ln ^{2}x}{(1+x)^{2}}dx &=\int _{0}^{1}\frac {\ln ^{2}x}{(1+x)^{2}}dx+\int _{1}^{\infty }\frac {\ln ^{2}x}{(1+x)^{2}}dx\\ &=2\int _{0}^{1}\frac {\ln ^{2}x}{(1+x)^{2}}dx\\ & =2\left [\frac {x}{1+x}\ln ^{2}x\right ]_0^1-2\int _{0}^{1}\frac {x}{1+x}\frac {2\ln x}xdx\\ &=-4\int _{0}^{1}\frac {\ln x}{1+x}dx\\ &=-4\left (-\frac {\pi ^{2}}{12}\right )\\ &=\textcolor {blue}{\frac {\pi ^{2}}{3}}. \end {aligned}$$

$$\begin {aligned} \int _{0}^{\infty }\frac {\ln ^{2}x}{(1+x^{2})^{2}}dx &=\int _{0}^{1}\frac {\ln ^{2}x}{(1+x^2)^{2}}dx+\int _{1}^{\infty }\frac {\ln ^{2}x}{(1+x^{2})^{2}}dx\\ &=\int _{0}^{1}\frac {\ln ^{2}x}{(1+x^{2})^{2}}dx+\int _{0}^{1}\frac {x^2\ln ^{2}x}{(1+x^{2})^{2}}dx\\ &=\int _{0}^{1}\frac {\ln ^{2}x}{1+x^{2}}dx\\ &=\Gamma (3)\beta (3)\\ &=\textcolor {blue}{\frac {\pi ^{3}}{16}}. \end {aligned}$$