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Question Number 142275 by mnjuly1970 last updated on 29/May/21

$$$$$$.......Advanced...\ast \ast \ast \ast \ast ...Integral......$$$$Provethat::\mathrm{\Phi }:={\int }_0^1\frac{1-x}{(1-x+x^2)log\left(x\right)}dx=$$$$proof::$$$$\mathrm{\Phi }:={\int }_0^1\frac{1-x^2}{(1-x^3)log\left(x\right)}dx$$$$f\left(a\right):={\int }_0^1\frac{1-x^a}{(1-x^3)log\left(x\right)}$$$$\mathrm{\Phi }:=f\left(2\right)........✓$$$$f{}^{\prime }\left(a\right):={\int }_0^1\frac{-x^{a}log\left(x\right)}{(1-x^3)log\left(x\right)}={\int }_0^1\frac{-x^a}{1-x^3}dx\left(★\right)$$$$\left(★\right)::x^3=y\Rightarrow f{}^{\prime }\left(a\right):=\frac{1}{3}{\int }_0^1\frac{y^{\frac{-2}{3}}-y^{\frac{a}{3}-\frac{2}{3}}}{1-y}dy$$$$:=\frac{1}{3}{\int }_0^1\frac{y^{\frac{-2}{3}}-1+1-y^{\frac{a}{3}-\frac{1}{3}}}{1-y}dy$$$$:=\frac{1}{3}(\psi (\frac{a}{3}+\frac{2}{3})-\psi \left(\frac{2}{3}\right))$$$$f\left(a\right):=log\left(\mathrm{\Gamma }(\frac{a}{3}+\frac{2}{3})\right)-\frac{a}{3}\psi \left(\frac{2}{3}\right)+C$$$$f\left(0\right):=0=log\left(\mathrm{\Gamma }\left(\frac{2}{3}\right)\right)+C$$$$C:=-log\left(\mathrm{\Gamma }\left(\frac{2}{3}\right)\right)$$$$\mathrm{\Phi }:=f\left(2\right)=log\left(\mathrm{\Gamma }\left(\frac{4}{3}\right)\right)-\frac{2}{3}\psi \left(\frac{2}{3}\right)-log\left(\mathrm{\Gamma }\left(\frac{2}{3}\right)\right)$$$$:=log\left(\frac{\mathrm{\Gamma }\left(\frac{4}{3}\right)}{\mathrm{\Gamma }\left(\frac{2}{3}\right)}\right)-\frac{2}{3}\psi \left(\frac{2}{3}\right)....✓$$$$$$

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