Closed form for $\int_0^1\frac{x}{\ln(x+1)(x^3+3x+3)}dx$ How can I evaluate the closed form of the following integral:
$$\int_0^1\frac{x}{\ln(x+1)(x^3+3x+3)}dx$$  

According to Wolfram Alpha, the numerical value of this integral is close to 0.2673, but it doesn't show up any closed form.

 A: I am almost sure that this integral does not have closed form solutions in terms of special functions. As a matter of fact even Feynmann would not be able to solve it as somebody else in here rightfully noted.
Yet if we were to modify the polynomial in the denominator so that it has nice roots then indeed we can come up with something. Consider the following integral instead:
\begin{eqnarray}
J:=\int\limits_0^1 \frac{x}{\log(1+x) (x^3+3 x^2+3 x+1)} dx =?
\end{eqnarray}
Now define a following function:
\begin{equation}
I(b) := \int\limits_0^1 \frac{x (1+x)^b}{\log(1+x) (x^3+3 x^2+3 x+1)} dx 
\end{equation}
subject to $I(-\infty)=0$. Then clearly $\left.I(b)\right|_{b=0} = J$. All we have to do now is to compute $I^{'}(b)$ and then integrate the result from minus infinity to zero. It turns out that the former can be done always, i.e. no matter what polynomial we have in the denominator whereas the later can be only done in very particular cases (meaning if the polynomial in question has nice roots). As usual by " can   only  be done in particular cases" I mean that I won't know  how to do it given my scarce mathematical knowledge.
Let us proceed:
\begin{eqnarray}
I^{'}(b) &=& \int\limits_0^1 \frac{x (1+x)^b}{ (x^3+3 x^2+3 x+1)} dx\\
&=& \int\limits_0^1 x (1+x)^{b-3} dx\\
&\underbrace{=}_{u:=1/(1+x)}& \int\limits_{1/2}^1 \frac{1-u}{u^b} d u\\
&=& \frac{4-3 \cdot 2^b+2^b b}{4(b-1)(b-2)}
\end{eqnarray}
Therefore the result reads:
\begin{eqnarray}
J = I(0) &=& \int\limits_{-\infty}^0 \frac{4-3 \cdot 2^b+2^b b}{4(b-1)(b-2)} db\\
&=& \log(2) + \frac{1}{4} \int\limits_0^\infty \exp(-\log(2) b)\left( \frac{1}{b+2} - \frac{2}{b+1}\right) db\\
&=& \log(2) - Ei(-2\log(2)) + Ei(-\log(2))
\end{eqnarray}
where $Ei()$ is the exponential integral function.
Now if we try to do the same procedure for the other integral in the question above we will end up with integrating hypergeometric function with respect to their parameters which is something they definitely have not taught me in my primary school so I wouldn't be able to get any result.
