Integral $\int_0^1\frac{\ln(1+x)}{1+x^2}\left(\frac{3\arctan x}{x}+2\ln x\right)dx$

Question

I was playing around with PARI GP to generate a challenging integral.

The method: The function lindep is intended to detect integer dependence.

The constants used: $\text{G}\ln 2,\pi^3,\pi\ln^2 2,\ln^3 2,\pi^2\ln 2,\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx$.

All these numbers are of "degree" $3$.

To get a challenging integral you multiply, $\ln(1+x),\ln(1+x^2),\dfrac{1}{1+x^2},\arctan x,\dfrac{1}{1+x},\dfrac{1}{x}$

The idea is to sum up two integrals to cancel out the awful term $\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx$

After ten minutes, I have found: \begin{align}A&=\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}\,dx\\ &=-3\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx+...\\ B&=\int_0^1\frac{\ln(1+x)\arctan x}{x(1+x^2)}\,dx\\ &=2\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx+...\\ \end{align} Now, if you take $2A+3B$ you cancel out the term $\int_0^{\frac{\pi}{4}}\ln^2(\cos x)\,dx$. The result is: \begin{align}\int_0^1\frac{\ln(1+x)}{1+x^2}\left(\frac{3\arctan x}{x}+2\ln x\right)dx=\frac{5}{128}\pi^3-\frac{7}{4}\text{G}\ln 2+\frac{3}{16}\pi \ln^2 2\end{align} The question is: How to prove this?

Addendum: My first candidate for $B$ was: $\int_0^1\frac{\ln(1+x)\arctan x}{1+x^2}\,dx$ but it's not expressible as an integer linear combination of the chosen constants. Probably not related to constants of "degree" $3$. To decrease the "degree" i have multiply by $\dfrac{1}{x}$ and voilà !

Answer 1

$$2A+3B=2\int_0^1 \frac{ \ln(1+x)\ln x}{1+x^2}dx+3 \int_0^1 \frac{ \arctan x\ln(1+x)}{x(1+x^2)}dx$$ $$=\color{blue}{2\underbrace{\int_0^1 \frac{ \ln(1+x)\ln x}{1+x^2}dx}_{=A}}+\color{green}{3\underbrace{\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}_{=C}}\color{red}{-3\underbrace{\int_0^1 \frac{x\arctan x \ln(1+x)}{1+x^2}dx}_{=D}}$$

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$$A\overset{IBP}=\underbrace{\arctan x\ln x\ln(1+x)\bigg|_0^1}_{=0}-\int_0^1 \frac{\arctan x\ln(1+x)}{x}dx-{\int_0^1 \frac{\arctan x \ln x}{1+x}dx}$$ $$=-C+\frac{\pi^3}{64}-\frac{G\ln 2}{2}$$ See here for above the integral.

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For $D$ let $x=\frac{1-t}{1+t}$ to get: $$D=\int_0^1 \frac{x\arctan x\ln(1+x)}{1+x^2}dx=\int_0^1 \left(\frac{\pi}{4}-\arctan t\right)\ln\left(\frac{2}{1+t}\right)\left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt$$ Sum up the two integral from above to get: $$2D=\frac{\pi}{4}\int_0^1 \ln\left(\frac{2}{1+t}\right)\left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt$$ $$-\ln 2\int_0^1 \arctan t \left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt$$ $$\require{cancel}+\int_0^1 \arctan t \ln(1+t)\left(\frac{1}{1+t}-\cancel{\frac{t}{1+t^2}}\right)dt$$ Here, I showed combined with your approach that: $$\boxed{\int_0^1\frac{\arctan x\ln(1+x)}{1+x}dx=\frac23 \underbrace{\int_0^1\frac{\arctan x\ln(1+x)}{x}dx}_{=C}-\frac{\pi^3}{128}+\frac{3\pi}{32}\ln^2 2}$$ And the first two integrals are pretty trivial. $$\frac{\pi}{4}\int_0^1 \ln\left(\frac{2}{1+t}\right)\left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt=\frac{\pi}{4}\int_0^1 \ln\left(1+\frac{1-t}{1+t}\right)\frac{1-t}{1+t} \frac{dt}{1+t^2}$$ $$=\frac{\pi}{4}\int_0^1 \frac{x\ln(1+x)}{1+x^2}dx=\frac{\pi^3}{384}+\frac{\pi}{32}\ln^2 2$$ See here. Similarly for the next integral: $$\ln 2\int_0^1 \arctan t \left(\frac{1}{1+t}-\frac{t}{1+t^2}\right)dt=\ln 2 \int_0^1 \frac{x\left(\frac{\pi}{4}-\arctan x\right)}{1+x^2}dx=\frac{\pi}{4}\ln^2 2-\frac{G}{2}\ln 2$$ $$\Rightarrow D=\boxed{-\frac{\pi^3}{384}-\frac{\pi}{16}\ln^2 2+\frac{G\ln 2}{4}+\frac13 C}$$

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Now we can breathe, because: $$2A+3B=\color{blue}{-\cancel{2C}+\frac{\pi^3}{32}-G\ln 2} \color{green}{+\cancel{3C}}\color{red}{+\frac{\pi^3}{128}+\frac{3\pi}{16}\ln^2 2-\frac{3G\ln 2}{4}-\cancel{C}}$$ $$\Rightarrow \boxed{2\int_0^1 \frac{ \ln(1+x)\ln x}{1+x^2}dx+3 \int_0^1 \frac{ \arctan x\ln(1+x)}{x(1+x^2)}dx=\frac{5\pi^3}{128}+\frac{3\pi}{16}\ln^2 2-\frac{7G\ln 2}{4}}$$