Any idea how to integrate this over the reals without using wolfram alpha?
Your welcome to go into $\Bbb{C}$ for the integration as long as its the same result over $\Bbb{R}$
I do recognize something that sort of looks like an arc tan in there $\int \frac{(x^2+1)} {(x^4+1)}$? any level of mathemagical trickery is fine.
On
Peter Foreman gave a nice trick from the classical analysis. If you are willing to use complex analysis (in case of definite integral from $-\infty$ to $\infty$), then it's easy to see, that the integrand has 2 roots of $x^4+1=0$ as simple poles in the upper half-plane: $x=(i\pm1)/\sqrt2$.
The residues can be easily calculated with the formula $g(z)/h'(z)$: $$ \mathrm{Res}\left(\frac{x^2+1}{x^4+1},\frac{i\pm1}{\sqrt2}\right) = -\frac{i}{2\sqrt2}. $$
Finally, the integral (we take the countour around half-circle, arc part will go to zero): $$ \int_{-\infty}^{\infty} \frac{x^2+1}{x^4+1} dx = 2\pi i\sum \mathrm Res = \pi\sqrt2 $$
Use $$\frac{x^2+1}{x^4+1}=\frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2}.$$