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.
$$
\int_{0}^{1}\frac{\mathrm{d}x}{\left(x + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}\,,
\qquad\qquad\int_{0}^{1}\frac{\mathrm{d}x}{\left(x^{2} + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}
$$
 A: 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
$$ I_a=\frac{\pi }{\sin(\pi/3)(1+a)^{1/3}}=\frac{2\pi}{\sqrt{3}(1+a)^{1/3}}\,.$$
A: Let's avoid complex numbers if we don't need it. First, try the substitution $x = 1/z$, so $dx = -dz/z^2$. This changes the lower and upper limits of integration to $\infty$ and $1$ respectively, which we can flip and cancel the minus sign introduced by the differential substitution. The integrand, after some simple algebra, simplifies to $\frac{z^2}{(z+1)(z-1)^{1/3}}$. Thus, the original integral is equivalent to $\int_1^\infty \frac{dz}{(z+1)(z-1)^{1/3}}$. 
From here, we can use another substitution $u = (z-1)^{1/3}$ which changes the integral to $\int_0^\infty \frac{3udu}{u^3+2}$. The denominator can be factored using a sum of cubes formula, then partial fractions and splitting the numerator will give 3 terms. One will simplify to $\arctan$ after substitutions, while the other two will simplify to $\log$. Plug in your endpoints from the final substitution (or backsub to get an explicit indefinite integral in $x$) and you should get the answer marty cohen got from WolframAlpha.
Though I haven't tried it, I'm sure a similar approach can unpack the second integral as well, again avoiding complex numbers.
If you would like to use contour integration (which might make the computation easier), you can probably do it from the "$u$" equation with a clever choice of contour, but again I haven't tried it so no guarantees.
A: I don't have time to say much about this right now, but:
Call these integrals respectively $I_1, I_2$. Further to AlexanderJ93's point, the substitution $x=\dfrac{1}{1+u^3}$ gives $I_n=\int_0^\infty\dfrac{3u(1+u^3)^{n-1}du}{1+(1+u^3)^n}$. If we want to go further, we can use the residue theorem. For $n=1$, the denominator $2+u^3$ has first-order zeroes $-2^{1/3}\omega^k$ for $0\le k\le 2,\,\omega:=\exp\dfrac{2\pi i}{3}$; similarly, for $n=2$ the denominator has first-order zeroes $2^{1/6}\exp\dfrac{\pm\pi i}{4}\omega^k$.
A: There are lots of different methods to calculate your integrals, 
the simplest way is partial fraction expansion [wikipedia]
(https://en.wikipedia.org/wiki/Partial_fraction_decomposition):
$
\frac{1}{(x+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1}{3}}}=\frac{x^{%
\frac{4}{3}}}{2\left( 1-x\right) ^{\frac{1}{3}}}+\frac{\left( 1-x\right) ^{%
\frac{2}{3}}}{x^{\frac{2}{3}}}+\frac{\left( 1-x\right) ^{\frac{2}{3}}x^{%
\frac{4}{3}}}{2\left( 1+x\right) }$
$\frac{1}{(x^{2}+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1}{3}}}=%
\frac{x^{\frac{4}{3}}}{2\left( 1-x\right) ^{\frac{1}{3}}}+\frac{\left(
1-x\right) ^{\frac{2}{3}}}{x^{\frac{2}{3}}}+\left( 1-x\right) ^{\frac{2}{3}%
}x^{\frac{1}{3}}-\frac{\left( 1-x\right) ^{\frac{2}{3}}x^{\frac{4}{3}}}{%
2\left( 1+x^{2}\right) }-\frac{\left( 1-x\right) ^{\frac{2}{3}}x^{\frac{7}{3}%
}}{2\left( 1+x^{2}\right) }$
or just the following transformation: 
$\int_{0}^{1}\frac{1}{(x+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1}{3}%
}}dx=\int_{0}^{1}\frac{x^{-\frac{2}{3}}}{(x+1)\left( 1-x\right) ^{\frac{1}{3}%
}}dx$
$\int_{0}^{1}\frac{1}{(x^{2}+1)\left( x^{2}\left( 1-x\right) \right) ^{\frac{1%
}{3}}}dx=\int_{0}^{1}\frac{x^{-\frac{2}{3}}}{(x^{2}+1)\left( 1-x\right) ^{%
\frac{1}{3}}}dx$
The following antiderivatives appear:
$\int \frac{x^{\alpha }}{\left( 1-x\right) ^{\beta }}=B\left( x,1+\alpha
,1-\beta \right) $
where B is the incomplete Beta -
Function.
$\int \frac{\left( 1-x\right) ^{\alpha }x^{\beta }}{\left( 1+x\right) }=\frac{%
x^{\alpha +1}}{\alpha +1}F_{1}\left( \alpha +1;\beta ,1;\alpha
+2;x,-x\right)$
where $F_{1}$ is the Appell hypergeometric function
AppellF1.
Moreover using the partial fraction decompositions [Dieckmann]
(http://www-elsa.physik.uni-bonn.de/dieckman/InfProd/InfProd.html):
$\frac{1}{\left( 1+x^{2}\right) }=\frac{i}{2\left( i+x\right) }-\frac{i}{%
2\left( x-i\right) }$
the last antiderivative terms are calculated by:
$\int \frac{\left( 1-x\right) ^{\alpha }x^{\beta }}{\left( 1+x^{2}\right) }=%
\frac{1}{2}\frac{x^{\alpha +1}}{\alpha +1}F_{1}\left( \alpha +1;\beta
,1;\alpha +2;x,i~x\right) +\frac{1}{2}\frac{x^{\alpha +1}}{\alpha +1}%
F_{1}\left( \alpha +1;\beta ,1;\alpha +2;x,-i~x\right) $
In special cases, the AppellF1 function can be reduced : 
$F_{1}\left( a;b_{1},b_{1};c;x,-x\right) =~_{3}F_{2}\left( \frac{a}{2}+\frac{1%
}{2},\frac{a}{2},b_{1};\frac{c}{2}+\frac{1}{2},\frac{c}{2};x^{2}\right) $
to a hypergeometric function
(http://functions.wolfram.com/HypergeometricFunctions/AppellF1/03/05/0002/)
Put in all the parameters and integration limits finally leads to
the mentioned results [marty cohen] (https://math.stackexchange.com/posts/2611282/edit). 
A: As I mentioned above, let's use complex analysis.  I think it is easier to do the second integral first.  To begin, consider the complex integral
$$\oint_Cdz \,  \frac{z^{-2/3} (z-1)^{-1/3}}{z^2+1}  $$
where $C$ is the following contour:

where the radius of the large circular arc is $R$ and the radius of the small circular arcs is $\epsilon$.
To evaluate the contour integral, we break the contour into various pieces and evaluate each piece by parametrizing accordingly.  In this case, the contour integral is equal to
$$e^{i \pi} \int_R^{\epsilon} dx \, \frac{e^{-i 2 \pi/3} x^{-2/3} e^{-i \pi/3} (x+1)^{-1/3}}{x^2+1} +i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\epsilon^{-2/3} e^{-i 2 \phi/3} \left ( \epsilon e^{i \phi} -1 \right)^{-1/3}}{\epsilon^2 e^{i 2 \phi}+1} \\ + \int_{\epsilon}^{1-\epsilon} dx \, \frac{x^{-2/3} e^{-i \pi/3} (1-x)^{-1/3}}{x^2+1} + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{\left (1+\epsilon e^{i \phi} \right)^{-2/3} \epsilon^{-1/3} e^{-i \phi/3}}{\left (1+\epsilon e^{i \phi}\right )^2+1} \\ + \int_{1-\epsilon}^{\epsilon} dx \, \frac{x^{-2/3} e^{i \pi/3} (1-x)^{-1/3}}{x^2+1} + i \epsilon \int_0^{-\pi} d\phi \, e^{i \phi} \frac{\epsilon^{-2/3} e^{-i 2 \phi/3} \left ( \epsilon e^{i \phi} -1 \right)^{-1/3}}{\epsilon^2 e^{i 2 \phi}+1} \\ +e^{-i \pi} \int_{\epsilon}^R dx \, \frac{e^{i 2 \pi/3} x^{-2/3} e^{i \pi/3} (x+1)^{-1/3}}{x^2+1} + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{R^{-2/3} e^{-i 2 \theta/3} \left ( R e^{i \theta}-1 \right )^{-1/3}}{R^2 e^{i 2 \theta}+1}$$
We consider the limits as $\epsilon \to 0$ and $R \to \infty$.  Independent of these limits, the first and seventh integrals cancel.  In these limits, the reader can verify that the second, fourth, sixth, and eighth integrals approach zero.  We are thus left with the third and fifth integrals.  The contour integral in the above limits then becomes
$$-i 2 \sin{\frac{\pi}{3}} \int_0^1 dx \, \frac{x^{-2/3} (1-x)^{-1/3}}{1+x^2}$$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles $z=\pm i = e^{\pm i\pi/2}$.  Note that the choice of the argument of the poles is not really a choice, as we have already defined this choice with our branch cut.  Accordingly, the contour integral is also equal to
$$i 2 \pi \left [\frac{e^{-i \pi/3} \left (e^{i \pi/2}-1 \right )^{-1/3} }{i 2} + \frac{e^{i \pi/3} \left (e^{-i \pi/2}-1 \right )^{-1/3} }{-i 2} \right ] $$
With the branch cut we have defined, note that $e^{i \pi/2}-1 = 2^{1/2} e^{i 3 \pi/4}$ and $e^{-i \pi/2}-1 = 2^{1/2} e^{-i 3 \pi/4}$.  Doing the rest of the arithmetic and setting the result equal to the above contour integral, we conclude that

$$\int_0^1 dx \, \frac{x^{-2/3} (1-x)^{-1/3}}{1+x^2} = 2^{-1/6} \pi \frac{\sin{\frac{5 \pi}{12}}}{\sin{\frac{\pi}{3}}} = \frac{\sqrt{2+\sqrt{3}}}{2^{1/6} \sqrt{3}} \pi$$

I leave it as an exercise for the reader to prove that the result is equal to that produced by Wolfram Alpha/Mathematica as shown by Robert Israel.
A: Here is an alternative evaluation for the second integral.
\begin{align*}
\int_0^1 \frac{dx}{(1 + x^2) [x^2 (1 - x)]^{1/3}} \, dx
&= \int_0^1 \sum_{n = 0}^\infty (-1)^n x^{2n} \frac{1}{x^{2/3} (1 - x)^{1/3}} \, dx\\
&= \sum_{n = 0}^\infty (-1)^n \int_0^1 x^{2n - 2/3} (1 - x)^{-1/3} \, dx\\
&= \sum_{n = 0}^\infty (-1)^n \text{B} \left (2n + \frac{1}{3}, \frac{2}{3} \right )\\
&= \sum_{n = 0}^\infty (-1)^n \frac{\Gamma (2n + 1/3) \Gamma (2/3)}{\Gamma (2n + 1)}\\
&= \Gamma \left (\frac{2}{3} \right ) \sum_{n = 0}^\infty \frac{(-1)^n \Gamma (2n + 1/3)}{(2n)!} \tag1
\end{align*}
Here $\text{B}(x,y)$ is Euler's Beta function while $\Gamma (x)$ is the familiar Gamma function. Now we just need to find a closed form for the sum appearing in (1). 
From the integral definition for the Gamma function, namely
$$\Gamma (z) = \int_0^\infty t^{z - 1} e^{-t} \, dt,$$
we can write
$$\Gamma \left (2n + \frac{1}{3} \right ) = \int_0^\infty t^{2n - 2/3} e^{-t} \, dt,$$
so the sum appearing in (1) can be rewritten as
\begin{align*}
\sum_{n = 0}^\infty \frac{(-1)^n \Gamma (2n + 1/3)}{(2n)!} &= \sum_{n = 0}^\infty \frac{(-1)^n}{(2n)!} \int_0^\infty \frac{t^{2n} e^{-t}}{t^{2/3}} \, dt\\
&= \int_0^\infty t^{-2/3} e^{-t} \left [\sum_{n = 0}^\infty \frac{(-1)^n t^{2n}}{(2n)!} \right ] \, dt\\
&= \int_0^\infty t^{-2/3} e^{-t} \cos t \, dt\\
&= \mathfrak{R} \int_0^\infty t^{-2/3} e^{-(1 - i) t} \, dt\\
&= \mathfrak{R} \left (\frac{1}{\sqrt[3]{1 - i}} \right ) \cdot \int_0^\infty u^{-2/3} e^{-u} \, du\\
&= \Gamma \left (\frac{1}{3} \right ) \mathfrak{R} \left (\frac{1}{\sqrt[3]{1 - i}} \right ).
\end{align*} 
Now it can be readily shown that
$$\frac{1}{\sqrt[3]{1 - i}} = \frac{1}{2^{1/6}} \exp \left (\frac{i \pi}{12} \right ).$$
Thus
$$\mathfrak{R} \left (\frac{1}{\sqrt[3]{1 - i}} \right ) = \frac{1}{2^{1/6}} \cos \left (\frac{\pi}{12} \right ) = \frac{1 + \sqrt{3}}{2 \cdot 2^{2/3}},$$
since $\cos (\pi/12) = (1 + \sqrt{3})/(2\sqrt{2})$, and we have
$$\sum_{n = 0}^\infty \frac{(-1)^n \Gamma (2n + 1/3)}{(2n)!} = \Gamma \left (\frac{1}{3} \right ) \frac{1 + \sqrt{3}}{2 \cdot 2^{2/3}}.$$
Substituting this result into (1), on taking advantage of the duplication formula for the Gamma function, namely
$$\Gamma \left (\frac{1}{3} \right ) \Gamma \left (\frac{2}{3} \right ) = \Gamma \left (\frac{1}{3} \right ) \Gamma \left (1 - \frac{1}{3} \right ) = \frac{\pi}{\sin (\pi/3)} = \frac{2\pi}{\sqrt{3}},$$
the value for the integral will be
$$\int_0^1 \frac{dx}{(1 + x^2) x^{2/3} \sqrt[3]{1 - x}} = \frac{\pi (1 + \sqrt{3})}{2^{2/3} \sqrt{3}}.$$
A: 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
$
