How to compute the integral $int_{0}^{infty} frac{x^{1/3}}{1+x^{2}} dx$

I want to compute this integral

$$\displaystyle\int_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}} \ dx$$

What I did was the following. I substituted $x=t^{6}$, so that my $dx= 6t^{5} \ dt$ and so the integral changes to

$$\int_{0}^{\infty} \frac{t^2}{1+t^{12}} \cdot 6 t^{5} \ dt =6 \cdot \int_{0}^{\infty} \frac{t^7}{1+t^{12}} \: dt$$

Now If I substitute $t^{4}=v$ then what I will be having is the following integral

$$\frac{6}{4}\cdot \int_{0}^{\infty} \frac{v}{1+v^{3}} \ dv$$

Now I can write $1+v^{3} = (1+v) \cdot (1-v+v^{2})$ and so I have

$$\begin{align*} \int_{0}^{\infty} \frac{v}{1+v^{3}}\: dv &= \int_{0}^{\infty}\biggl[\frac{1}{3}\cdot \frac{v+1}{1-v+v^{2}} - \frac{1}{3} \cdot \frac{1}{1+v}\biggr]\: dv \end{align*}$$

Now the point is that the integral of $1/(1+v) \to \infty$, so I am not sure if this is the right way to do. Can anyone suggest anything?

Answer

Substitute $x^{1/3} = t$, i.e., $x = t^3$, i.e., $dx = 3t^2 dt$. Hence,

$$I = \int_0^{\infty} \dfrac{t}{1+t^6} 3t^2 dt = 3 \int_0^{\infty} \dfrac{t^3}{1+t^6}dt$$
$$\int_0^{\infty} \dfrac{t^3}{1+t^6}dt = \int_0^{1} \dfrac{t^3}{1+t^6}dt + \int_1^{\infty} \dfrac{t^3}{1+t^6}dt$$
$$\int_1^{\infty} \dfrac{t^3}{1+t^6}dt = \int_1^0 \dfrac{1/t^3}{1+1/t^6}\left(-\dfrac{dt}{t^2}\right) = \int_0^1 \dfrac{t}{t^6+1} dt$$

Hence,
$$\dfrac{I}3 = \int_0^1 \dfrac{t+t^3}{1+t^6} dt = \int_0^1 \dfrac{t}{1-t^2+t^4} dt = \dfrac1{2 \sqrt{3}} \left(\int_0^1 \dfrac{dt}{t^2 - \sqrt3 t + 1} - \int_0^1 \dfrac{dt}{t^2 + \sqrt3 t + 1} \right)$$
I trust you can finish it off from here.