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Wolfram alpha show that the following integral :$\displaystyle\int_{0}^{1}\frac{\ln²(x)(2x-x²-2ln(x+1))}{x^3(1-x)}dx $ dosn't converge but really it's converge and have this result :

$$\displaystyle\int_{0}^{1}\frac{\ln²(x)(2x-x²-2ln(x+1))}{x^3(1-x)}dx=\frac{5}{4}\zeta(4)-7\ln2\zeta(4)+\frac{7}{2}\zeta(3)-\frac{3}{2}\zeta(2)+8\ln2-0.25$$. Edit I have edit the question because i have a wrong typo place from the paper Really the above integral has been showed by ANTHONY SOFO in him interesting paper,page25

My question here is :Why the titled integral result has not been added to wolfram alpha for giving the correct computations ?

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    Why do you write $\ln^2(x+1)\ln(x+1)$ instead of $\ln^3 (x+1)$? – 2017-01-05
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    As written, it diverges at $1$. – 2017-01-05
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    pleas check the paper which published in sarajevo journal by Anthony soft – 2017-01-05
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    The integral in the paper (Remark 1) has $\ln^2x$, not $\ln^2(x+1)$. WolframAlpha gives a finite value for the corrected integral. Voting to close. – 2017-01-05
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    sorry for that , i have going to edit the for – 2017-01-05
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    As we try again: https://www.wolframalpha.com/input/?i=integral(+%5Cfrac%7Bln%C2%B2(x)ln(x%2B1)%7D%7Bx(1-x)%7D)dx,+x%3D0to+1 – 2017-01-05
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    FWIW, independent of the typo I think the tone of the question is also poor. 'Why hasn't Wolfram Alpha done this thing?' is going to be innately opinionated; 'Why hasn't Alpha added this very narrow integral?' sounds quite strongly like an accusation to me. – 2017-01-05
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    Yeah, WolframAlpha is not magic. It has many limitations... for example, it could not solve the following sum when $x\in\mathbb N$:$$\sum_{n=0}^\infty\frac1{(n^2+1)^x}$$ – 2017-01-05
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    @simpleArt well I do not feel so bad that I can not solve the above if WolframAplha can not either! ;) – 2017-01-05

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Integral diverges (as a Lebesgue or Riemann integral), because if $x\rightarrow1$, function is $\sim\frac{ln(2)^{3}}{1-x}$