I was messing around in Desmos and noticed that $\frac{1}{\tan^{-1}\left(x^{2}\right)+1}$ had an asymptote at $y=\frac{2}{\pi+2}$. I wondered what $\int_{-\infty}^{\infty}\left(\frac{1}{\tan^{-1}\left(x^{2}\right)+1}-\frac{2}{\pi+2}\right)dx$ would be (assuming it converges) and Desmos gave me $1.14450701513$. However, I want an exact answer (in a closed form preferably) so I also tried to solve it analytically:
$$\int_{-\infty}^{\infty}\left(\frac{1}{\tan^{-1}\left(x^{2}\right)+1}-\frac{2}{\pi+2}\right)dx$$
$$2\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{u+1}-\frac{2}{\pi+2}\right)\frac{x^4+1}{2x}du$$
$$\frac{1}{\pi+2}\int_{0}^{\frac{\pi}{2}}\frac{\left(\pi-2u\right)\sec^{2}u}{\left(u+1\right)\sqrt{\tan u}}du$$
After integration by parts with help from @KStarGamer, I got:
$$2\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\tan u}}{\left(u+1\right)^{2}}du$$
Wolfram Alpha says I can represent the integrand as:
$$x^\frac{1}{2}-2x^\frac{3}{2}+O(x^\frac{5}{2})$$
But I have not yet learned series expansions or big O notation so I may have made a mistake and I don't know where to go from here.