[tex]\displaystyle \int_{0}^1\dfrac{1+x^2}{1+x^2+x^4}\,dx =\int_{0}^1\dfrac{1+x^2}{1+x^2+x^4}\, dx = \int_{0}^1\dfrac{x^{-2}}{x^{-2}}\cdot \dfrac{1+x^2}{1+x^2+x^4}}\, dx = \\ \\ = \int_{0}^1\dfrac{x^{-2}+1}{x^{2}+x^{-2}+1}\, dx = \int_{0}^1\dfrac{x^{-2}+1}{x^2+x^{-2}-2+3}\, dx = \\ \\ =\int_{0}^1\dfrac{x^{-2}+1}{(x-x^{-1})^2+3}\,dx=[/tex]
[tex]\displaystyle x-x^{-1} = t \Rightarrow (1+x^{-2})\,dx = dt \\ x = 0 \Rightarrow t \to -\infty \\ x = 1 \Rightarrow t = 0 \\ \\ =\int_{-\infty}^1\dfrac{1}{t^2+(\sqrt 3)^2}\, dt = \dfrac{1}{\sqrt 3}\,\text{arctg}\,\Big(\dfrac{t}{\sqrt 3}\Big)\Big|_{-\infty}^0 = 0-\dfrac{1}{\sqrt 3}\cdot \Big(-\dfrac{\pi}{2}\Big) = \\ \\ \\= \boxed{\dfrac{\pi}{2\sqrt 3}}[/tex]
