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

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
$$\int_0^1 \frac{dx}{(1 + x^2) x^{2/3} \sqrt[3]{1 - x}} = \frac{\pi (1 + \sqrt{3})}{2^{2/3} \sqrt{3}}.$$
