Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$

I think you can do it this way.

\begin{align*} \int \frac{1}{x^4 +1} \ dx & = \frac{1}{2} \cdot \int\frac{2}{1+x^{4}} \ dx \\\ &= \frac{1}{2} \cdot \int\frac{(1-x^{2}) + (1+x^{2})}{1+x^{4}} \ dx \\\ &=\frac{1}{2} \cdot \int \frac{1-x^2}{1+x^{4}} \ dx + \frac{1}{2} \int \frac{1+x^{2}}{1+x^{4}} \ dx \\\ &= \frac{1}{2} \cdot -\int \frac{1-\frac{1}{x^2}}{\Bigl(x+\frac{1}{x})^{2} - 2} \ dx + \text{same trick} \end{align*}

$$\int\frac 1{1+x^4}dx=\frac12\int\frac{1+x^2+1-x^2}{1+x^4}dx$$

$$\int\frac{1+x^2}{1+x^4}dx=\int\frac{\frac1{x^2}+1}{\left(x-\frac1x\right)^2+2}dx$$

Set $x-\frac1x=\sqrt2\tan\phi$

$$\int\frac{1-x^2}{1+x^4}dx=-\int\frac{1-\frac1{x^2}}{\left(x+\frac1x\right)^2-2}dx$$

Set $x+\frac1x=\sqrt2\sec\psi$

Reference: Trigonometric substitution

HINT : $$x^4+1 =(x^2+1)^2-2x^2 =(x^2+\sqrt 2x+1)(x^2-\sqrt2x+1)$$