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?

As an alternative approach:$$I=\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx\,\overset{\large x^{2/3}=u}=\,\frac32 \int_0^\infty \frac{1}{u^3+1}du$$
We will substitute $\displaystyle{u=\frac{1-t}{1+t}\Rightarrow du=-\frac{2}{(1+t)^2}dt}$. 
The reason behind it is that $(1-t)^3+(1+t)^3=2(3t^2+1)$, thus we get rid of the third powers.
$$\Rightarrow I=\frac32 \int_{-1}^1 \frac{t+1}{3t^2+1}dt=\frac32 \cdot 2\int_0^1 \frac{dt}{3t^2+1}dt=3\cdot \frac{1}{\sqrt 3}\arctan(\sqrt 3 t)\bigg|_0^1=\frac{\pi}{\sqrt 3}$$
