$\int\limits_{}^{}\frac{1}{x^{4}+1}dx=\int\limits_{}^{}\frac{1}{(x^{2}-\sqrt{2}x+1)(x^{2}+\sqrt{2}x+1)}dx$$=\frac{1}{2\sqrt{2}}\int\limits_{}^{}(\frac{\sqrt{2}-x}{x^{2}-\sqrt{2}x+1}+\frac{\sqrt{2}+x}{x^{2}+\sqrt{2}x+1})dx$$=\frac{1}{2\sqrt{2}}\int\limits_{}^{}(\frac{\sqrt{2}-x}{(x-\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}}+\frac{\sqrt{2}+x}{(x+\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}})dx$$A=\int\limits_{}^{}\frac{\sqrt{2}-x}{(x-\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}}dx$Đặt $u=x-\frac{1}{\sqrt{2}}$$A=\int\limits_{}^{}\frac{\sqrt{2}-2u}{1+2u^{2}}du$$B=\int\limits_{}^{}\frac{\sqrt{2}+x}{(x+\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}}dx$Đặt $v=x+\frac{1}{\sqrt{2}}$$B=\int\limits_{}^{}\frac{\sqrt{2}+2v}{1+2v^{2}}dv$

$\int\limits_{}^{}\frac{1}{x^{4}+1}dx=\int\limits_{}^{}\frac{1}{(x^{2}-\sqrt{2}x+1)(x^{2}+\sqrt{2}x+1)}dx$$=\frac{1}{2\sqrt{2}}\int\limits_{}^{}(\frac{\sqrt{2}-x}{x^{2}-\sqrt{2}x+1}+\frac{\sqrt{2}+x}{x^{2}+\sqrt{2}x+1})dx$$=\frac{1}{2\sqrt{2}}\int\limits_{}^{}(\frac{\sqrt{2}-x}{(x-\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}}+\frac{\sqrt{2}+x}{(x+\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}})dx$$A=\int\limits_{}^{}\frac{\sqrt{2}-x}{(x-\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}}dx$Đặt $u=x-\frac{1}{\sqrt{2}}$$A=\int\limits_{}^{}\frac{\sqrt{2}-2u}{1+2u^{2}}du=\int\limits_{}^{}\frac{\sqrt{2}}{1+2u^{2}}-\frac{2u}{1+2u^{2}} $$B=\int\limits_{}^{}\frac{\sqrt{2}+x}{(x+\frac{1}{\sqrt{2}})^{2}+(\frac{1}{\sqrt{2}})^{2}}dx$Đặt $v=x+\frac{1}{\sqrt{2}}$$ B=\int\limits_{}^{}\frac{\sqrt{2}+2v}{1+2v^{2}}dv=\int\limits_{}^{}\frac{\sqrt{2}}{1+2v^{2}}+\frac{2v}{1+2v^{2}}$