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?
 A: 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.
$$\Rightarrow I=\frac32 \int_{-1}^1 \frac{t+1}{3t^2+1}dt=\frac32 \cdot 2\int_0^1 \frac{dt}{3t^2+1}dt=3\cdot \frac{1}{\sqrt 3}\arctan(\sqrt 3 t)\bigg|_0^1=\frac{\pi}{\sqrt 3}$$
A: 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?
A: 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 
$$\int_0^\infty \frac{\mathrm dx}{x^{1/3}(1+x^2)}=\frac12\mathrm{B}(1/3,2/3)$$
A: Quickest way would be to let $x\mapsto\tan x$ and use the beta function$$\begin{align*}\mathfrak{I} & =\int\limits_0^{\pi/2}\mathrm dx\,\tan^{-1/3}x\\ & =\int\limits_0^{\pi/2}\mathrm dx\,\sin^{-1/3}x\cos^{1/3}x\\ & =\frac 12\operatorname{B}\left(\frac 13,\frac 23\right)\end{align*}$$A direct application of Euler’s Reflection formula gives$$\int\limits_0^{\infty}\frac {\mathrm dx}{x^{1/3}(1+x^2)}\color{blue}{=\frac {\pi}{\sqrt3}}$$
A: 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}
