Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?

HINT:
Putting $x=\frac1y,dx=-\frac{dy}{y^2}$
$$I=\int_0^\infty\frac{dx}{1+x^4}=\int_\infty^0\frac{-dy}{y^2\left(1+\frac1{y^4}\right)}$$
$$=-\int_\infty^0\frac{y^2dy}{1+y^4}=\int_0^\infty\frac{y^2dy}{1+y^4} \text{ as } \int_a^bf(x)dx=-\int_b^af(x)dx$$
$$I=\int_0^\infty\frac{y^2dy}{1+y^4}=\int_0^\infty\frac{x^2dx}{1+x^4}$$
$$\implies 2I=\int_0^\infty\frac{dx}{1+x^4}+\int_0^\infty\frac{x^2dx}{1+x^4}=\int_0^\infty\frac{1+x^2}{1+x^4}dx=\int_0^\infty\frac{\frac1{x^2}+1}{\frac1{x^2}+x^2} dx$$
Now the idea is to express the denominator as a polynomial of $\int \left(\frac1{x^2}+1 \right)dx =x-\frac1x=u\text{(say)}$
$$\text{The denominator =}\frac1{x^2}+x^2=\left(x-\frac1x\right)^2+2=u^2+2$$
Now, complete the definite integral with $u$
