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This integral can be found in Cornel's book, (Almost) Impossible Integral, Sums and Series page $97$ where he showed that

$$\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt=\frac14\left(\frac12\ln^2(1+x^2)-2\operatorname{Li}_2(x)+\frac12\operatorname{Li}_2(-x^2)+\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)\right)$$

by differentiating $\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)$ then integrating back. How magical is that?

My approach: \begin{align} I&=\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt\overset{IBP}{=}\frac12\ln(1+x^2)\ln(1-x)+\frac12\int_0^x\frac{\ln(1+t^2)}{1-t}\ dt\\ &\overset{1-x=u}{=}\frac12\ln(1+x^2)\ln(1-x)-\frac12\int_{1}^{1-x}\frac{\ln(2+2u+u^2)}{u}\ du\\ &=\frac12\ln(1+x^2)\ln(1-x)-\Re\int_{1}^{1-x}\frac{\ln((1+i)-u)}{u}\ du \end{align} and by using $\displaystyle\int \frac{\ln(a-x)}{x}\ dx=\ln(a)\ln x-\operatorname{Li}_2\left(\frac{x}{a}\right)\ $, we get \begin{align} I&=\frac12\ln(1+x^2)\ln(1-x)-\Re\left(\ln(1+i)\ln u-\operatorname{Li}_2\left(\frac{u}{1+i}\right)\right)_{u=1}^{u=1-x}\\ &=\frac12\ln(1+x^2)\ln(1-x)-\Re\left(\ln(1+i)\ln(1-x)-\operatorname{Li}_2\left(\frac{1-x}{1+i}\right)+\operatorname{Li}_2\left(\frac{1}{1+i}\right)\right)\\ &=\frac12\ln(1+x^2)\ln(1-x)+\frac12\ln2\ln(1-x)+\Re\operatorname{Li}_2\left(\frac{1-x}{1+i}\right)-\frac{\pi^2}{16}\\ &\boxed{=\frac12\ln\left(2(1+x^2)\right)\ln(1-x)+\Re\operatorname{Li}_2\left(\frac{1-x}{1+i}\right)-\frac{\pi^2}{16}} \end{align} where I used $\ \Re\ln(1+i)=\frac12\ln2\ $ and $\ \Re\operatorname{Li}_2\left(\frac{1}{1+i}\right)=\frac{\pi^2}{16}$.

I tested my solution numerically and all is good but as we can see it does not match Cornel's answer. any idea how to make it match?

Thanks,

Its worth to mention that by comparing the two results proved above, we find the new interesting identity:

$$\Re\operatorname{Li}_2\left(\frac{1-x}{1+i}\right) =\frac14\left(\frac12\ln^2(1+x^2)-2\ln\left(2(1+x^2)\right)\ln(1-x)-2\operatorname{Li}_2(x)+\frac12\operatorname{Li}_2(-x^2)\\+\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)+\frac{\pi^2}{4}\right)$$

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    $\begingroup$ @ automaticallyGenerated yes because they both give the same results. the question is how to make them match. $\endgroup$
    – Ali Shadhar
    Commented Jul 5, 2019 at 20:05
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    $\begingroup$ Are you interested in other approaches? Because I don't have much knowldege about dilogarithms in order to match that. (Not saying that I have one now, but I'm thinking of an approach).// Btw, maybe you find this interesting: $$\int_0^x \frac{t\ln(1+t)}{1+t^2}dt=\frac12 \operatorname{Li}_{2}\left(\frac{1-x^2}{2}\right)+\frac12\ln(1-x^2)\ln\left(\frac{1+x^2}{2}\right)-\frac12 \operatorname{Li}_2\left(\frac12\right) -\frac14\left(\frac12\ln^2(1+x^2)-2\operatorname{Li}_2(x)+\frac12\operatorname{Li}_2(-x^2)+\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)\right)$$ $\endgroup$
    – Zacky
    Commented Jul 5, 2019 at 21:04
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    $\begingroup$ Commenting since it doesn't work. My idea was to consider the following integrals:$$I=\int_0^x \frac{t\ln(1-t)}{1+t^2}dt, \quad J=\int_0^x \frac{t\ln(1+t)}{1+t^2}dt$$ $$I+J=\int_0^x\frac{t\ln(1-t^2)}{1+t^2}dt\overset{t^2=y}=\frac12\int_0^{x^2}\frac{\ln(1-y)}{1+y}dy$$ $$=\frac12 \left(\operatorname{Li}_{2}\left(\frac{1-x^2}{2}\right)+\ln(1-x^2)\ln\left(\frac{1+x^2}{2}\right)- \operatorname{Li}_2\left(\frac12\right)\right)$$ Similarly for $I-J$, just let $\frac{1-t}{1+t}$ and use partial fractions. Then $2I=(I+J)+(I-J)$. From the above the integral result that I mentioned easily follows. $\endgroup$
    – Zacky
    Commented Jul 5, 2019 at 21:49
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    $\begingroup$ @Zacky $x$ can be any number. if we use my result or yours for $x=1$, it does not seem to work but its fine if we use Cornels'. now we can see the big difference between these results. $\endgroup$
    – Ali Shadhar
    Commented Jul 5, 2019 at 21:50
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    $\begingroup$ nice idea Zacky. thats one of my fav techniques. $\endgroup$
    – Ali Shadhar
    Commented Jul 5, 2019 at 21:52

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