$\int \frac{x^2-1}{x^4+1}dx=$

$$\begin{gathered} I=\int \frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^4+1}\mathrm{dx}=\int \frac{1-\frac{1}{{\mathrm{x}}^2}}{{\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}}\mathrm{dx} \\ put x+\frac{1}{x}=t\Rightarrow \left(1-\frac{1}{x^2}\right)dx=dt \\ and t^2=x^2+\frac{1}{x^2}+2 \Rightarrow t^2-2=x^2+\frac{1}{x^2} \end{gathered}$$

$\begin{array}{l}I=\int \frac{\mathrm{dt}}{{\mathrm{t}}^2-2}\\ =\frac{1}{2\sqrt{2}}\log \left|\frac{\mathrm{t}-\sqrt{2}}{\mathrm{t}+\sqrt{2}}\right|+\mathrm{c}\\ =\frac{1}{2\sqrt{2}}\log \left|\frac{\mathrm{x}+\frac{1}{\mathrm{x}}-\sqrt{2}}{\mathrm{x}+\frac{1}{\mathrm{x}}+\sqrt{2}}\right|+\mathrm{c}\end{array}$

$=\frac{1}{2\sqrt{2}}\log \left|\frac{{\mathrm{x}}^2-\sqrt{2}\mathrm{x}+1}{{\mathrm{x}}^2+\sqrt{2}\mathrm{x}+1}\right|+\mathrm{c}$