<s>[INST] How to compute $\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$ 
How to compute
$$\int_0^{+\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt 2}.$$
  I'm interested in more ways of computing this integral.

There is always the straight forward method to decompose into simple elements 
$\dfrac{1}{(1 + t^{4})}$ it works but it is tedious. If someone has a faster and clever method, I'm interested :)
Update :


*

*My quesion is different than this question because all the solution that state there is talk about the the straight forward method to decompose into simple elements $\dfrac{1}{(1 + t^{4})}$ as i said before i'm not intersted in that way

 [/INST] Here is another method using purely clever substitutions. Note that 
$$
\int_0^\infty \frac{\mathrm{d}x}{1+x^4}
=
\int_{0}^\infty \frac{w^2}{1+w^4} \mathrm{d}w
$$
Where the last integral comes from the substitution $w \mapsto 1/x$. Addition now gives that
$$
\begin{align*}
\int_0^\infty \frac{\mathrm{d}x}{1+x^4}
& = \frac{1}{2}\int_0^\infty \frac{1+x^2}{1+x^4}\mathrm{d}x \\
& = \frac{1}{2}\int_0^\infty \frac{1+1/x^2}{x^2+1/x^2} \mathrm{d}x \\
& = \frac{1}{2}\int_0^\infty \frac{(x-1/x)'}{(x-1/x)^2+2} \mathrm{d}x \\
& = \frac{1}{2}\int_{-\infty}^\infty \frac{\mathrm{d}u}{u^2+2} 
  = \frac{1}{2\sqrt{2}}\int_{-\infty}^\infty \frac{\mathrm{d}y}{1+y^2}
  = \frac{\pi}{2\sqrt{2}}
\end{align*}
$$
Where the substitutions $u \mapsto x - 1/x$ and $y \mapsto \sqrt{2} u$ were used.
</s>