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How can we find the value of $$\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$$ using elementary methods?

With some help of calculator I get the result: $\displaystyle{\frac3{128}\pi^3-\frac9{32}\pi\ln^22}$.

Thoughts of this integral
Since I have asked this question and Pisco gave a brilliant answer, I tried to convert $$I_1=\int_0^1\arctan x\ln(1+x)\frac{dx}x\text{ and }I_2=\int_0^1\arctan x\ln(1+x)\frac{dx}{1+x}$$ into the form of integral Pisco gave.
Integrating by parts to the second integral converts $I_2$ into $\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx$.
But for $I_1$? Integrating by parts gives a dilog function and I tried substitution $x=\frac{1-t}{1+t}$ and got $$\frac{\ln\frac{2}{t+1} \arctan\frac{1-t}{1+t}}{1-t^2}$$ which is not what I want.

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    $\begingroup$ Just curious, but why do you expect there to be an elementary method? $\endgroup$ – Number Dec 28 '18 at 10:15
  • $\begingroup$ @Zacky I expect there to be an elementary method since I think the answer of the linked question is quite elementary. $\endgroup$ – Kemono Chen Dec 28 '18 at 10:23
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Here is an elementary approach, although it turned into a crossover with FDP's answer.

First note that from here we have: $$\color{blue}{\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}=\frac{3}{2}\int_0^1 \frac{\arctan x\ln(1+x^2)}{x}dx$$ $$\overset{IBP}=\frac32 \underbrace{\ln x\arctan x\ln(1+x^2)\bigg|_0^1}_{=0}-\frac32 \left(\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx+2\int_0^1 \frac{x\arctan x\ln x}{1+x^2}dx\right) $$ Back to the original integral, we have: $$I=\color{blue}{2\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}-\color{red}{3\int_0^1 \frac{\arctan x \ln(1+x)}{1+x}dx} $$ $$=\color{blue}{-3\left(\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx+2\int_0^1 \frac{x\arctan x\ln x}{1+x^2}dx\right)}-\color{red}{3\int_0^1\frac{\arctan x\ln(1+x)}{1+x}dx}$$ $$\Rightarrow I=-3(B+2A+J)\quad \quad (1)$$ Where I kept the notation like in FDP's answer. Namely: $$\begin{align*} \displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\ \displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\ \displaystyle J&=\int_0^1\dfrac{\arctan x\ln(1+x)}{1+x}dx \end{align*}$$ Also another two identities follows from that post, see $(8)$ and $(9)$: $$J=\dfrac{5}{3}G\ln 2-\dfrac{\pi^3}{128}+\dfrac{3\pi\left(\ln 2\right)^2}{32}+B+\dfrac{2}{3}\left(\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}\right)-\dfrac{2}{3}\cdot\frac{\pi^3}{32} $$ $$\Rightarrow \color{purple}{J=2G\ln 2 -\frac{5\pi^3}{128}+\frac{3\pi}{32}\ln^2 2 +B} \tag 2$$ $$\color{magenta}{A=\dfrac{1}{64}\pi^3-B-G\ln 2} \tag 3$$ Now plugging $(2)$ and $(3)$ in $(1)$ yields: $$I=-3\left(B+2\left(\color{magenta}{\dfrac{1}{64}\pi^3-B-G\ln 2}\right)+ \color{purple}{2G\ln 2 -\frac{5\pi^3}{128}+\frac{3\pi}{32}\ln^2 2 +B}\right)$$ $$\Rightarrow I=-3\left(-\frac{\pi^3}{128}+\frac{3\pi}{32}\ln^2 2\right)=\boxed{\frac{3\pi^3}{128}-\frac{9\pi}{32}\ln^2 2}$$ Credits to FDP for his amazing answer there!

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NOT A FULL SOLUTION BUT A START:

Here you have:

\begin{equation} I = \int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx = \int_0^1\arctan x\ln(1+x)\left[\frac{2 - x}{x(x + 1)}\right]dx \end{equation}

Consider using Feynman's Trick with two parameters:

\begin{equation} I(a,b) = \int_0^1\arctan(ax)\ln(1+bx)\left[\frac{2 - x}{x(x + 1)}\right]dx \end{equation}

Here $I = I(1,1)$ and $I(0,b), I(a,0) = 0$. Here take the partial derivative with respect to $a$ and $b$ to yield:

\begin{equation} \frac{\partial^2I}{\partial a \partial b} = \int_0^1\frac{x}{a^2x^2 + 1}\cdot\frac{x}{1 +bx}\left[\frac{2 - x}{x(x + 1)}\right]dx = \int_0^1 \frac{x\left(2 - x\right)}{\left(a^2x^2 + 1\right)\left(1 + bx\right)\left(x + 1\right)}dx \end{equation}

From here employ Partial Fraction Decomposition. I will finish off in an hour if you're still interested (sorry will be afk for the next hour).

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  • $\begingroup$ Yes, you are correct! thanks for that. I will edit now. $\endgroup$ – user150203 Dec 28 '18 at 11:31
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    $\begingroup$ I have a feeling that it might be better to represent $\arctan(x)$ and/or $\ln(x)$ as an integral and covert the original single integral to a double/triple. Will have to investigate tomorrow as I'm falling asleep as I type this. Great Question btw. $\endgroup$ – user150203 Dec 28 '18 at 11:39
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For $I_1$, by integration by parts, \begin{eqnarray*} I_1&=&\int_0^1\arctan x\ln(1+x)d\ln x\\ &=&\arctan x\ln(1+x)\ln x|_0^1-\int_0^1\ln x\left(\frac{\ln(1+x)}{1+x^2}+\frac{\arctan x}{1+x}\right)dx\\ &=&-\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-\int_0^1\frac{\ln x\arctan x}{1+x}dx\\ &=&-I_3-I_4. \end{eqnarray*} Here $I_3$ and $I_4$ are $$ I_3=\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx, I_4=\int_0^1\frac{\ln x\arctan x}{1+x}dx. $$ From here, $$ I_3= -2 G \ln (2)-3 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{11 \pi ^3}{128}+\frac{3}{32} \pi \ln ^2(2). $$ From here, $$ I_4=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64} $$ $G$ is the Catalan's constant.

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