Evaluate: $\displaystyle\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+1 } dx }$ equals ;

$\displaystyle \int \dfrac{x^2 + 1}{x^4 + 1} . dx$

$\displaystyle \int \dfrac{1 + x^{-2}}{x^2 + x^{-2}}. dx$ [divide with $x^2$]

$\displaystyle \int \dfrac{1 + x^{-2}}{x^2 + x^{-2} -2 + 2} .dx$

$\displaystyle \int \dfrac{1 + x^{-2}}{(x - x^{-1})^2 + 2} dx$

$u = x - x^{-1}$

$\displaystyle \int \dfrac{dU}{U^2 + 2} = \dfrac{1}{\sqrt{2}} \tan^{-1} \dfrac{U}{\sqrt{2}} + C$

$= \dfrac{1}{\sqrt{2}} \tan^{-1} \dfrac{(x - x^{-1})}{\sqrt{2}} + C$

$= \dfrac{1}{\sqrt{2}} \tan^{-1} \left(\dfrac{x^2 - 1}{\sqrt{2} x} \right) + C$