I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx$$ With no success, i tried to consider the following integrals $$I=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)}{1+x^2}\:dx,J=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx$$ $$I+J=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^2\right)}{1+x^2}\:dx=\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x^4\right)}{1+x^2}\:dx-\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$$ I managed to express that $1$st integral into somewhat known euler sums but that $2$nd integral arrived at a sum i didnt know how to evaluate which was $$2\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{\left(2k+1\right)^3}$$ And it seems this approach wont go smooth, could i tackle the main integral differently? maybe with an easier approach?
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stefan
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