∫\(\frac{x^2+1}{x^4+1}\)dx=

\(\frac{1}{\sqrt {2 }}\)tan^-1 \((\frac{x^2-1}{\sqrt{2}x}\))+C

$I=∫\frac{x^2+1}{x^4+1}dx$

$I=∫\frac{x^2(1+\frac{1}{x^2})}{x^2(x^2+\frac{1}{x^2})}dx$

$\Rightarrow I=∫\frac{(1+\frac{1}{x^2})}{(x-\frac{1}{x})^2+2x×\frac{1}{x}}⇒∫\frac{(1+\frac{1}{x^2})}{(x-\frac{1}{x})^2+2}dx$

Let $t=x-\frac{1}{x}⇒dt=(1+\frac{1}{x^2})dx$  ...(i)

$I=∫\frac{dt}{t^2+2}⇒∫\frac{dt}{t^2+(\sqrt{2})^2}=\frac{1}{\sqrt{2}}tan^{-1}(\frac{t}{\sqrt{2}})+C$

$\frac{1}{\sqrt{2}}tan^{-1}(\frac{(x-\frac{1}{x})}{\sqrt{2}})+C .....$ where $t = x-\frac{1}{x}$

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