Compute $\int_0^{\infty} \frac {1}{x^{1/3}(1+x^2)}dx$

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. $$\Rightarrow I=\frac32 \int_{-1}^1 \frac{t+1}{3t^2+1}dt=\frac32 \cdot 2\int_0^1 \frac{dt}{3t^2+1}dt=3\cdot \frac{1}{\sqrt 3}\arctan(\sqrt 3 t)\bigg|_0^1=\frac{\pi}{\sqrt 3}$$

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?

Also using the Beta function, but in a different way.

Recall that $$\mathrm{B}(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt$$ Using the substitution $w=\frac{1-t}t$, we see that $$\mathrm{B}(a,b)=\int_0^\infty\frac{w^{b-1}}{(1+w)^{a+b}}\mathrm dw$$ Then using the sub $w=u^2$, $$\mathrm{B}(a,b)=2\int_0^\infty\frac{u^{2b-1}}{(1+u^2)^{a+b}}\mathrm du$$ So setting $2b-1=-1/3$, and $a+b=1$, we see that $$\int_0^\infty \frac{\mathrm dx}{x^{1/3}(1+x^2)}=\frac12\mathrm{B}(1/3,2/3)$$