# How to calculate $I=\int_0^{\infty} \frac{dx}{x^4+a^4}$ [duplicate]

I understand it is an even function, which indicates $$I=\frac{1}{2}\int_{-\infty}^{\infty} \frac{dx}{x^4+a^4}$$

What should I do in the next step?

• you can divide everything by $a^4$ to assume, essentially, $a=1$. Then you can either (1) use complex analysis or (2) factor the denominator into $(x^2-\sqrt{2}x-1)(x^2+\sqrt{2}x-1)$, use partial fractional decomposition, then integrate from there (using one or two other tricks such as completing the square). – mathworker21 Aug 18 '19 at 14:08
• I'm sure you can find both steps done by googling "integral of 1/(x^4+1)" – mathworker21 Aug 18 '19 at 14:08

Writing $$\frac{1}{a^4\left(\left(\frac{x}{a}\right)^4+1\right)}$$ now substitute $$t=\frac{x}{a}$$ so $$dt=\frac{1}{a}dx$$ and factorize $$t^4+1=t^4+2t^2+1-2t^2=(t^2+1)^2-2t^2=(t^2+1-\sqrt{2}t)(t^2+1+\sqrt{2}t)$$