indefinite integrals

\hspace{-16}\int\frac{x^4}{x^4+1}dx=\int1.dx-\frac{1}{x^4+1}dx\\\\\\ x-\frac{1}{2}.\int\frac{(x^2+1)-(x^2-1)}{x^4+1}dx$\\\\\\ $x-\frac{1}{2}.\int\frac{x^2+1}{x^4+1}dx+\frac{1}{2}.\int\frac{x^2-1}{x^4+1}dx$\\\\\\ Let $I_{1}=\frac{1}{2}.\int\frac{x^2+1}{x^4+1}dx$ and $I_{2}=\frac{1}{2}.\int\frac{x^2-1}{x^4+1}dx$\\\\\\ Now Calculate $I_{1}=\frac{1}{2}.\int\frac{x^2+1}{x^4+1}dx$\\\\\\ $=\frac{1}{2}.\int\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=\frac{1}{2}.\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+(\sqrt{2})^2}dx$\\\\\\ Now Let $x-\frac{1}{x}=t\Leftrightarrow (1+\frac{1}{x^2})dx=dt$\\\\\\ $\frac{1}{2}.\int\frac{1}{t^2+(\sqrt{2})^2}dx=\frac{1}{2.\sqrt{2}}.\tan^{-1}\left(\frac{t}{\sqrt{2}}\right)=\frac{1}{2.\sqrt{2}}.\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)$

$Now Calculate $I_{1}=\frac{1}{2}.\int\frac{x^2-1}{x^4+1}dx$\\\\\\ $=\frac{1}{2}.\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=\frac{1}{2}.\int\frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2-(\sqrt{2})^2}dx$\\\\\\ Now Let $x+\frac{1}{x}=a\Leftrightarrow (1-\frac{1}{x^2})dx=da$\\\\\\ So $\frac{1}{2}.\int\frac{1}{a^2-(\sqrt{2})^2}dx=\frac{1}{2.\sqrt{2}}.\ln\left|\frac{a-\sqrt{2}}{a+\sqrt{2}}\right|$\\\\\\ $\frac{1}{2.\sqrt{2}}.\ln\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|$\\\\\\ So $\int\frac{x^4}{1+x^4}dx=x-\frac{1}{2.\sqrt{2}}.\ln\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+\frac{1}{2.\sqrt{2}}.\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C$