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I tried to solve this integral and got it, I showed firstly $$\int_0^1 \frac{\ln^2(x+1)+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx=2\Im\left[\text{Li}_3(1+i) \right] $$ and for other integral $$\int_0^1 \frac{\ln\left(\frac{2x}{x^2+1}\right)\ln x}{x^2+1} dx=\int_0^1 \frac{\left(\ln2+\ln x\right)\ln x}{x^2+1} dx-\int_0^1 \frac{\ln\left(x^2+1\right)\ln x}{x^2+1} dx$$ then by using series and its value $$ \int_0^1 \frac{\left(\ln2+\ln x\right)\ln x}{x^2+1} dx=-G\ln2+\frac{\pi^3}{16}$$ $$ \int_0^1 \frac{\ln\left(x^2+1\right)\ln x}{x^2+1} dx=\sum_{k=1}^{\infty}(-1)^{n-1}H_n \int_0^1x^{2n}\ln x dx $$ $$=\sum_{k=1}^{\infty} \frac{(-1)^{n} H_n}{(2n+1)^2} =2\Im\left[\text{Li}_3(1+i) \right]+\frac{3}{32}\pi^3+\frac{\pi}{8}(\ln2)^2-G\ln2$$ then got the result $$\int_0^1 \frac{\ln^2(x+1)-\ln\left(\frac{2x}{x^2+1}\right)\ln x+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx=\frac{\pi}{8} \left(\frac{\pi^2}{4}+(\ln2)^2\right)$$ MY QUESTION

How can I get the result with simple way without using complicated constants and values of some series which are long to prove it?

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    $\begingroup$ Plan of work (leading to a perfect solution): One way to approach such a problem is to reduce the main integral to arctan-log integrals that will be further calculated together or separately. In the case of sums of arctan-log integrals, there is a rich spectrum of (useful) results both on this site and in Cornel's books. Proceeding this way, you'll avoid the use of results with constants like polylogarithmic values involving a complex argument (and you can also totally avoid the use of series). I'll give you some references to consider and build yourself a solution. (continued) $\endgroup$
    – user97357329
    Commented Oct 1, 2023 at 15:21
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    $\begingroup$ So, after integrating by parts once, you might find it useful to check: math.stackexchange.com/questions/1842284; math.stackexchange.com/q/3050696; math.stackexchange.com/q/3647097; math.stackexchange.com/q/3054741; (Almost) Impossible Integrals, Sums, and Series (2019), Section 1.26, page 17; More (Almost) Impossible Integrals, Sums, and Series (2023), Sections $1.36$-$1.38$, pages $48$-$51$. All you need at this point is to cleverly employ such results and get your integral evaluated as you desire. (continued) $\endgroup$
    – user97357329
    Commented Oct 1, 2023 at 15:32
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    $\begingroup$ It is best for you to get used to these known results since you might want to return to them when meeting other integrals like the one you proposed where such reductions to arctan-log integrals are possible. If there is a more magical, faster way, I cannot tell you now, but the way I propose already offers you a lot of magic because of its high level of elegance. Arranging the integrals and employing the needed results is a boring part for me, but you might enjoy it! End of story $\endgroup$
    – user97357329
    Commented Oct 1, 2023 at 15:37
  • $\begingroup$ I just tried to solve the integral in the second line which solved using series and harmonic series $\endgroup$
    – Faoler
    Commented Oct 3, 2023 at 5:06

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