# 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.

• Do you know if the integral is supposed to have a nice closed form? – b00n heT May 14 '16 at 9:27
• Anything is possible. Also I am expecting a closed form including some special functions. – The Integral May 14 '16 at 9:40
• @TheIntegral, did you just come up with this integral at random? – Yuriy S May 14 '16 at 9:41
• @YuriyS Would you consider this as a random integral? link – The Integral May 14 '16 at 9:46
• @TheIntegral, what does 'random integral' even mean, and how is this related? I asked specifically if this integral is something you came up with, and what reasons do you have to expect a closed form? The polynomial in the denominator doesn't even have nice roots – Yuriy S May 14 '16 at 10:00

$$\begin{eqnarray} J:=\int\limits_0^1 \frac{x}{\log(1+x) (x^3+3 x^2+3 x+1)} dx =? \end{eqnarray}$$ Now define a following function: $$$$I(b) := \int\limits_0^1 \frac{x (1+x)^b}{\log(1+x) (x^3+3 x^2+3 x+1)} dx$$$$ subject to $$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.