How do I integrate these functions? I deleted my last question, since some of you wanted me to rewrite the question properly. I feel sorry for inconvenience, but please understand that this is the first time I use $\texttt{MathJax}$.

Up to now, I tried every method I know in integration, like substituition, partial fractions, uv-method, etc. But seems like nothing works. I would appreciate to have your help. Thanks.

$$
\int_{0}^{1}\frac{\mathrm{d}x}{\left(x + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}\,,
\qquad\qquad\int_{0}^{1}\frac{\mathrm{d}x}{\left(x^{2} + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}
$$
First integral can be rewrite in the following form:
$$I=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(x+1\right)x}\:dx$$
This view suggests the following change of variabel $u=x/(1-x)$, $dx=\dfrac{du}{(1+u)^2}$
$$I=\int _0^\infty\frac{ u^{-2/3}du}{(1+2u)}=\frac{1}{2^{1/3}}\int _0^\infty\frac{ u^{-2/3}du}{(1+u)}$$
The latter integral is the Beta function
$$\frac{1}{2^{1/3}}B(1/3,2/3)=\frac{1}{2^{1/3}} \Gamma(1/3)\Gamma(2/3)=\frac{2^{2/3}\pi}{\sqrt{3}}\,.$$
Here I use the identity $\Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\sin(\pi x)}$

To evaluate the secon integral I decompose the fraction $\dfrac{1}{1+x^2}=\frac{1}{2}\left(\dfrac{1}{1+i x}+\dfrac{1}{1-ix}\right)$
The integral 
$$I_a=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(1+a x\right)x}\:dx=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}$$
The integral $I_a$ is calculated absolutely analogously to the integral $I$.
Note that this identity is valid not only for real positive $a$ but also for complex one. It is related wit the fact that the integral 

$$I=\int _0^\infty\frac{ u^{-2/3}du}{(1+(1+a)u)}$$ can be evaluated for complex a using contour integration technique, but here I don't want touch it.

Thus we have 
$$ I_2=\int _0^1\frac{(x/(1-x))^{(1/3)}}{\left(1+x^2\right)x}\:dx=\frac{2\pi}{\sqrt{3}}\frac{1}{2}\left(\frac{1}{(1-i)^{1/3}} + \frac{1}{(1+i)^{1/3}}\right)=\frac{2\pi}{\sqrt{3}} \frac{cos(\pi/12)}{2^{1/6}}=\frac{\pi}{2^{2/3}} \frac{\sqrt3+1}{\sqrt3}
$$

EDIT:
Let us show that 
$$I_a=\int _0^\infty\frac{ u^{-2/3}du}{(1+(1+a)u)}=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}\,.$$
The integrant has a cut in the complex plane from zero to infinity. We set this cut along posinive real axis.
We will consider integral 
$$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}\,,$$
contour $\gamma$ is consists of two parts:

1) $z$ from $-i \epsilon +\infty$ to $-i\epsilon$ (against positive real axis a littel bit below cut

2) $z=\epsilon e^{-i \phi}$ $\phi$ from $\phi=\pi/2$ to $\phi=3\pi/2$ (semicircle)

3) $z$ from $i\epsilon$ to $i \epsilon +\infty$ (along positive real axis a littel bit higer cut

$$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}=I_a(1-e^{2\pi i/3 })$$ 
In the other side this function has a pole in the poin $z=-1/(1+a)$ and we can calculate this integral by residue
$$\int_\gamma \frac{ z^{-2/3}dz}{(1+(1+a)z)}=\frac{-2\pi i e^{i \pi/3}}{(1+a)^{1/3}}\,. $$ 
Thus we obtain
$$ I_a=\frac{\pi }{\sin(\pi/3)(1+a)^{1/3}}=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}\,.$$
Comment, but
easier to enter as an answer.

Wolfy says:

$\int_0^1 \dfrac{dx}{(x + 1) (x^2 (1 - x))^{1/3}}
 = \dfrac{2^{2/3} π}{\sqrt{3}}
≈2.87923
$

and

$\int _0^1\dfrac{dx}{(x^2+1)(x^2(1-x))^{1/3}}
= \dfrac{(3 + \sqrt{3}) π}{3\cdot 2^{2/3}}
≈3.1217
$