<s>[INST] 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.
$$
\int_{0}^{1}\frac{\mathrm{d}x}{\left(x + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}\,,
\qquad\qquad\int_{0}^{1}\frac{\mathrm{d}x}{\left(x^{2} + 1\right)
\left[x^{2}\left(1 - x\right)\right]^{1/3}}
$$
 [/INST] 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). 
</s>