<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] Hint. Taking from your last step,
$$\int\frac x{x^2+x+1}dx=\frac{1}{2}\int\frac{D(x^2+x+1)}{x^2+x+1}dx-\frac{1}{2}\int\frac{1}{(x+1/2)^2+3/4}dx.$$
Can you take it from here?
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