<s>[INST] Compute $\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$ Compute $$I=\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$$
My attempt:
$$u=x^{1/3}\implies I = 3\int_0^{\infty}\frac {u}{u^6+1}du=\frac 32\int_0^\infty \frac {1}{u^3+1}du=\frac 32 \int_0^\infty \frac {1}{(x+1)(x^2+x+1)}dx\\\implies I=\frac 32 \int_0^{\infty} \frac 1{x+1}-\frac x{x^2+x+1}dx$$ And I'm stuck here, what can I do from here?
 [/INST] Yet another approach. 
\begin{align}
\int_0^\infty \frac{dx}{\sqrt[3]{x} (1 + x^2)} &\overset{\large x \mapsto x^{3/2}}=\frac{3}{2} \int_0^\infty \frac{dx}{1 + x^3}\\
&= \frac{3}{2} \int_0^1 \frac{dx}{1 + x^3} + \frac{3}{2} \int_1^\infty \frac{dx}{1 + x^3}.
\end{align}
Enforcing a substitution of $x \mapsto 1/x$ in the second of the integrals leads to
\begin{align}
\int_0^\infty \frac{dx}{\sqrt[3]{x} (1 + x^2)} &= \frac{3}{2} \int_0^1 \frac{1 + x}{1 + x^3} \, dx\\
&= \frac{3}{2} \int_0^1 \frac{dx}{x^2 - x + 1}\\
&= \frac{3}{2} \int_0^1 \frac{dx}{\left (x - \frac{1}{2} \right )^2 + \frac{3}{4}}\\
&= \sqrt{3} \left [\tan^{-1} \left (\frac{2x - 1}{\sqrt{3}} \right ) \right ]_0^1\\
&= \frac{\pi}{\sqrt{3}}.
\end{align}
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