Integral

Note that $\frac{3x^2}{4x^4 + 5x^2 + 2} \; = \; \frac{(4x^2+1)-(x^2+1)}{1 + (4x^2+1)(x^2+1)}$ and hence the integral is $\begin{aligned} I & = \; \int_0^\infty \big[\tan^{-1}(4x^2+1) - \tan^{-1}(x^2+1)\big]\,\frac{dx}{x} \; = \; \lim_{X \to \infty , t \to 0} \int_t^X \big[\tan^{-1}(4x^2+1) - \tan^{-1}(x^2+1)\big]\,\frac{dx}{x}\\ & = \; \lim_{X \to \infty,t \to 0}\left(\int_{2t}^{2X} \tan^{-1}(x^2+1)\frac{dx}{x} - \int_t^X \tan^{-1}(x^2+1)\,\frac{dx}{x}\right) \; = \; \lim_{X \to \infty}\int_X^{2X}\tan^{-1}(x^2+1)\,\frac{dx}{x} - \lim_{t \to 0}\int_t^{2t} \tan^{-1}(x^2+1)\,\frac{dx}{x} \\ & = \; \lim_{X\to \infty}\int_{1/(2X)}^{1/X} \tan^{-1}(1 + x^{-2}) \frac{dx}{x} - \lim_{t \to 0} \int_t^{2t}\tan^{-1}(x^2+1)\,\frac{dx}{x} \; = \; \lim_{t \to 0}\int_t^{2t} \tan^{-1}(1 + x^{-2}) \frac{dx}{x} - \lim_{t \to 0} \int_t^{2t}\tan^{-1}(x^2+1)\,\frac{dx}{x}\\ & = \; \tfrac12\pi \ln2 - \tfrac14\pi \ln 2 \; = \; \boxed{\tfrac14\pi \ln2} \end{aligned}$