Integrate $f(x)=\frac{1}{1+(x-1/x)^2}$ I came across this rather simple looking function
$$
f(x)=\frac{1}{1+\left(x-\frac{1}{x}\right)^2}
$$
which has a nice looking plot with two peaks symmetric around the vertical axis, whereas each peak is not symmetric by itself. It also turns out to be difficult to integrate by hand, so I have two questions about it:

*

*What kind of function is it? does it have a name?

*How to actually integrate it? For example, if I know how the function behaves for small and large $x$, i.e.:

for very large $x$:
$$
\int\frac{1}{1+x^2} dx = \arctan(x) + C
$$
and for $x\sim 0$:
$$
\int\frac{1}{1+1/x^2} dx = x-\arctan(x) + C
$$
would it be possible to use this information to somehow get to them (at least one-sided) integral of the original function?

$$I=\int \frac{dx}{1+(x-1/x)^2}=\int\frac{x^2}{x^4-x^2+1}dx =\frac{1}{2} \int \left( \frac{1-\frac{1}{x^2}}{x^2+1/x^2-1} + \frac{1+\frac{1}{x^2}}{x^2+1/x^2-1}\right)dx$$
$$\implies I(x)=\frac{1}{2} \int \left( \frac{1-\frac{1}{x^2}}{(x+1/x)^2-3} + \frac{1+\frac{1}{x^2}}{(x-1/x)^2+1}\right)dx$$
Let $x+1/x=u$ and $x-1/x=v$ in the first and second integrals, respectivelt, then
$$I(x)=\frac{1}{2}\left (\int \frac{du}{u^2-3}+ \int \frac{dv}{v^2+1}\right).$$
$$I(x)=\frac{1}{2} \left( \frac{1}{2\sqrt{3}} \ln \frac{u-\sqrt{3}}{u+\sqrt{3}}+ \tan^{-1}v \right) +C.$$
