$\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx=$? Consider the convergent integral (I looked up difficult indefinite integrals on google images and then I saw this integrand and I was like hey let's see if it converges)
$$I=\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$$
We have the numerical approximation $$I\approx0.5553603672697931...$$ 
Which Wolfram alpha says is close to $\frac\pi{4\sqrt{2}}$. All my attempts at this integral have been fruitless and I need some help. Here's the only attempt of mine that actually made the integrand smaller:
$x=\tan u$: $$I=\int_{\pi/4}^{\pi/2}\frac{\tan^2u-1}{\sqrt{\tan^4u+1}}du$$
Next step:
bang head on floor in agony
Any suggestions?

$$
\begin{aligned}
\int_{1}^{\infty} \frac{x^{2}-1}{\left(x^{2}+1\right) \sqrt{x^{4}+1}} d x =& \int_{1}^{\infty} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right) \sqrt{x^{2}+\frac{1}{x^{2}}}} d x \\
=& \int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^{2}-2}} \\
=& \int_{2}^{\infty} \frac{d u}{u \sqrt{u^{2}-2}},\textrm{ where }u=x+\frac{1}{x}.  \\
=& \int_{2}^{\infty} \frac{1}{u^{2}} d\left(\sqrt{u^{2}-2}\right) \\
=& \int_{2}^{\infty} \frac{d \sqrt{u^{2}-2}}{\left(\sqrt{u^{2}-2}\right)^{2}+2} \\
=& \frac{1}{\sqrt{2}}\left[\tan ^{-1} \frac{\sqrt{u^{2}-2}}{\sqrt{2}}\right]_{2}^{\infty} \\
=& \frac{1}{\sqrt{2}}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)\\
=& \frac{\pi}{8} \sqrt{2}
\end{aligned}
$$
