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# Ahmed Integral doesn't evaluate

Posted 6 months ago
 Why doesn't Mathematica evaluate the well-known Ahmed integral? $$\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\left(x^2+1\right) \sqrt{x^2+2}} \, dx$$
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Posted 6 months ago
 Gracias Sorin Suciu. Ya conocía https://mathworld.wolfram.com/AhmedsIntegral.html. Mathematica puede evaluar formalmente integrales como: Integrate[ ArcTan[Sqrt[1 + x^2]]/(1 + x^2)^(3/2), {x, 0, 1}] // FullSimplify, Integrate[ArcTan[x]/((1 + x^2) x), {x, 1, Infinity}] , etc. Me sorprende que no calcule formalmente la muy conocida integral de Ahmed. Siguiendo el artículo " Integral de Ahmed: la solución inaugural " y con ayuda de Mathematica se obtiene:Integrate[(Pi/2)/(Sqrt[2 + x^2] (1 + x^2)), {x, 0, 1}] = [Pi]^2/12Apart[1/((1 + x^2) (2 + x^2 + y^2)), x] = 1/((1 + x^2) (1 + y^2)) - 1/((1 + y^2) (2 + x^2 + y^2))......etcIntegrate[ArcTan[Sqrt[2 + x^2]]/(Sqrt[2 + x^2] (1 + x^2)), {x, 0, 1}]=(5 [Pi]^2)/96
Posted 6 months ago
 How to solve see here and hereRegards M.I
Posted 6 months ago
 Gracias Sorin Suciu. Ya conocía https://mathworld.wolfram.com/AhmedsIntegral.html. Mathematica puede evaluar formalmente integrales como: Integrate[ ArcTan[Sqrt[1 + x^2]]/(1 + x^2)^(3/2), {x, 0, 1}] // FullSimplify, Integrate[ArcTan[x]/((1 + x^2) x), {x, 1, Infinity}] , etc. Me sorprende que no calcule formalmente la muy conocida integral de Ahmed. Siguiendo el artículo " Integral de Ahmed: la solución inaugural " y con ayuda de Mathematica se obtiene: Integrate[(Pi/2)/(Sqrt[2 + x^2] (1 + x^2)), {x, 0, 1}] = \[Pi]^2/12 Apart[1/((1 + x^2) (2 + x^2 + y^2)), x] = 1/((1 + x^2) (1 + y^2)) - 1/((1 + y^2) (2 + x^2 + y^2))......etc Integrate[ArcTan[Sqrt[2 + x^2]]/(Sqrt[2 + x^2] (1 + x^2)), {x, 0, 1}]=(5 \[Pi]^2)/96
Posted 6 months ago
 Mathematica did it this way:
Posted 6 months ago
 Looking at https://mathworld.wolfram.com/AhmedsIntegral.html it seems that Mathematica evaluates the generic integrals but not to the same results from the article.
Posted 6 months ago
 I will venture the obvious reason, that it does not know how to do this integral.