If got this question from here: Find a number $\alpha > 1$ such that the following holds
I was wondering if something with my answer is wrong, as it still has a downvote.
Let $\alpha$ be the golden ratio, so especially $\alpha=\frac{1}{\alpha-1}$ holds.
$\int\dfrac{dx}{(1+x^\alpha)^\alpha}=\int \dfrac{dx}{(1+x^\frac{1}{\alpha-1})^\alpha}=x(x^{\frac{1}{\alpha-1}}+1)^{1-\alpha}=x(x^\alpha+1)^{1-\alpha}$
So $\int^\infty_0\dfrac{dx}{(1+x^\alpha)^\alpha}=\lim_{x\rightarrow\infty} x(x^\alpha+1)^{1-\alpha}=\lim_{x\rightarrow\infty} \frac{x}{(x^\alpha+1)^\frac{1}{\alpha}}=1$