1708\hspace{-16}\bf{\int\frac{x^2+1}{x^4+1}dx}$\\\\\\ Like Rishab says......\\\\\\ Divide Both $\bf{N_{r}}$ and $\bf{D_{r}}$ By $\bf{x^2}$, We Get\\\\ $\bf{\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=\int\frac{\left(1+\frac{1}{x^2}\right)}{(x-\frac{1}{x})^2+(\sqrt{2})^2}dx}$\\\\\\ Now Put $\bf{x-\frac{1}{x}=t\Leftrightarrow \left(1+\frac{1}{x^2}\right)dx=dt}$\\\\\\ $\bf{\int \frac{1}{t^2+(\sqrt{2})^2}dt=\frac{1}{\sqrt{2}}.\tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)+C}$\\\\\\ So $\bf{\int\frac{x^2+1}{x^4+1}dx = \frac{1}{\sqrt{2}}.\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C}$
