A closed form of $\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx$ Is it possible to evaluate the following integral in a closed form?
$$\int_0^\infty\frac{\sqrt[\phi]{x}\ \arctan x}{\left(x^\phi+1\right)^2}dx,$$
where $\phi$ is the golden ratio:
$$\phi=\frac{1+\sqrt{5}}{2}.$$
 A: Since $\frac1\phi=\phi-1$, we get
$$
\begin{align}
\int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x
&=\int_0^\infty\frac{x^\phi\arctan(x)}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{1}\\
&=\int_0^\infty\frac{x^\phi(\frac\pi2-\arctan(x))}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{2}
\end{align}
$$
Average $(1)$ and $(2)$ to get
$$
\begin{align}
\int_0^\infty\frac{\sqrt[\phi]{x}\,\arctan(x)}{\left(x^\phi+1\right)^2}\mathrm{d}x
&=\frac\pi4\int_0^\infty\frac{x^\phi}{\left(x^\phi+1\right)^2}\frac{\mathrm{d}x}{x}\tag{3}\\
&=\frac\pi{4\phi}\int_0^\infty\frac{x}{\left(x+1\right)^2}\frac{\mathrm{d}x}{x}\tag{4}\\
&=\frac\pi{4\phi}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $\frac1\phi=\phi-1$
$(2)$: Substitute $x\mapsto\frac1x$
$(3)$: Average $(1)$ and $(2)$
$(4)$: Substitute $x\mapsto x^{1/\phi}$
$(5)$: $\int_0^\infty\frac{\mathrm{d}x}{(x+1)^2}=\left[-\frac1{x+1}\right]_0^\infty=1$
A: robjohn's result can be generalized to all real $a\ne0$:
$$\int_0^\infty\frac{x^{a-1}\arctan x}{(x^a+1)^2}dx=\frac\pi{4\,|a|}.$$
