A double sum or a definite integral.

Solution 1:

$$\int\limits_0^1\mathrm dx\,\frac {\log(1+x^3)}{1+x}\color{blue}{=\frac 12\operatorname{Li}_2\left(-\frac 13\right)+\frac 14\log^23+\frac 12\log^22-\frac {\pi^2}{36}}$$Confirmed by Wolfram Alpha.

Call the integral $\mathfrak{I}$ and make the transformation $x\mapsto\tfrac {1-x}{1+x}$. What's left is$$\mathfrak{I}=\log^22+\int\limits_0^1\mathrm dx\,\frac {\log(1+3x^2)}{1+x}-3\int\limits_0^1\mathrm dx\,\frac {\log(1+x)}{1+x}$$ The last integral is trivially equal to$$\int\limits_0^1\mathrm dx\,\frac {\log(1+x)}{1+x}\color{red}{=\frac 12\log^22}$$The middle integral is difficult and I am still not aware of a full solution to the integral by hand. Replace the three with a parameter say, $a$, and differentiate with respect to $a$$$\mathfrak{I}(a)=\int\limits_0^1\mathrm dx\space\frac {\log(1+ax^2)}{1+x}$$Thus$$\begin{align*}\mathfrak{I}'(a) & =\int\limits_0^1\mathrm dx\space\left[\frac 1{(1+a)(1+x)}+\frac x{(1+a)(1+ax^2)}-\frac 1{(1+a)(1+ax^2)}\right]\\ & =\frac 1{1+a}\log 2+\frac {\log(1+a)}{2a(1+a)}-\frac 1{\sqrt a(1+a)}\arctan\sqrt a\end{align*}$$Integrate both sides with respect to $a$ and from zero to three to get (I used Wolfram Alpha on this step)$$\int\limits_0^1\mathrm dx\space\frac {\log(1+3x^2)}{1+x}\color{brown}{=\log^22+\frac 12\operatorname{Li}_2\left(-\frac 13\right)+\frac 14\log^23-\frac {\pi^2}{36}}$$Now add everything together and you should get the result I stated at the beginning of my answer.

Solution 2:

If it helps: Maple 7 computes the expression $$ - \mathrm{ln}(2)\,\mathrm{ln}(3) - \mathrm{ln}(2)\,\mathrm{ ln}({\displaystyle \frac {1}{3}} \,I) - \mathrm{ln}(2)\,\mathrm{ ln}({\displaystyle \frac {-1}{3}} \,I) \\- \mathrm{dilog}( - {\displaystyle \frac {1}{3 + I\,\sqrt{3}}} + {\displaystyle \frac {I\,\sqrt{3}}{3 + I\,\sqrt{3}}} ) \\ - \mathrm{dilog}({\displaystyle \frac {1}{ - 3 + I\, \sqrt{3}}} + {\displaystyle \frac {I\,\sqrt{3}}{ - 3 + I\,\sqrt{ 3}}} ) \\ + \mathrm{dilog}( - {\displaystyle \frac {1}{ - 3 + I\, \sqrt{3}}} + {\displaystyle \frac {I\,\sqrt{3}}{ - 3 + I\,\sqrt{ 3}}} ) \\ + \mathrm{dilog}({\displaystyle \frac {1}{3 + I\,\sqrt{3} }} + {\displaystyle \frac {I\,\sqrt{3}}{3 + I\,\sqrt{3}}} ) $$

This will evaluate to $-0.12693500084879648110964091818$ and is compatible with Wolfram Alpha's $-0.126935$ (no symbolic answer here).

Please note that Maple's dilog is related to the standard polylogarithms by $$\operatorname{dilog(x)} = \operatorname{Li}_2(1-x)$$

Solution 3:

This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $\operatorname {Li_2}(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:

$$\int_0^1 \frac{ (1-2y)\ln(1+y)}{1-y+y^2} dy = -2\int_0^1 \frac{ (y-\frac12)\ln(1+y)}{(y-\frac12)^2 +\frac34} dy.$$

Then apply the substitution $x=y-\frac12$.

$$\int_{-\frac 12}^{\frac 12} \frac{x \ln(x+ \frac32)}{x^2+ \frac 34}dx = \int_{-\frac 12}^{\frac 12} \overbrace{\ln(x+ \frac32)}^{u} \cdot \underbrace {\frac {x}{x^2+ \frac 34} dx }_{dv}$$ Integration by parts show the original integral is equivalent to $\displaystyle \int_{-\frac 12}^{\frac12} \frac{\ln(x^2+\frac34)}{x+\frac 32}dx$. Now we can introduce yet another substitution, say $z=x+\frac 32$. Now we get:

$$\int_1^2 \frac {\ln \bigl((z- \frac 32)^2 + \frac 34\big)}{z}dz$$ We can factor the sum of squares by using the fact that $a^2 +b^2 = (a+bi)(a-bi)$. $$\int_1^2 \frac {\ln\big((z- \frac 32 + \frac {\sqrt 3}{2}i)(z- \frac 32 - \frac {\sqrt 3}{2}i)\big)}zdz$$

Using the log rules, you can rewrite this as the sum of two integrals: $$\int_1^2 \frac {\ln(z- \frac 32 + \frac {\sqrt 3}{2}i)}zdz + \int_1^2 \frac {\ln (z- \frac 32 - \frac {\sqrt 3}{2}i)}zdz \tag{1}$$

Now lets look for a general solution to $\displaystyle \int_1^2 \frac {\ln (x+a)}{x}dx $, where $a$ is any constant. $$\int_1^2 \frac {\ln (x+a)}{x}dx = \int_1^2 \frac {\ln\big(a(\frac xa +1)\big)}{x}dx = \ln(a) \int_1^2 \frac {dx}x+ \int_1^2 \frac {\ln(\frac xa +1)}{x} dx.$$ Apply the substitution $u = - \frac xa$. This changes it to:

$$ \ln(a)\ln(2)+ \int \frac {\ln(1-u)}{u} du = \Big[\ln(a)\ln(x)+ \operatorname {Li_2} (- \frac xa)+ C\Big]_1^2$$ $$ \int_1^2 \frac {\ln (x+a)}{x}dx =\ln(a)\ln(2)+ \operatorname {Li_2} (- \frac 2a)- \operatorname {Li_2} (-\frac 1a) \tag{2}$$

Now, putting (2) into (1) yields:

$$ \ln(- \frac 32 + \frac {\sqrt 3}{2}i)\ln(2)+ \operatorname {Li_2} (- \frac 2{- \frac 32 + \frac {\sqrt 3}{2}i})- \operatorname {Li_2} (-\frac 1{- \frac 32 + \frac {\sqrt 3}{2}i}) + \ln(- \frac 32 - \frac {\sqrt 3}{2}i)\ln(2)+ \operatorname {Li_2} ( \frac 2{ \frac 32 + \frac {\sqrt 3}{2}i})- \operatorname {Li_2} (\frac 1{ \frac 32 + \frac {\sqrt 3}{2}i}) $$

There's probably a way to simplify this, but I bet it's very tedious. The imaginary part should end up being 0.