Ahmed Integral doesn't evaluate

Questions
1

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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2

I will venture the obvious reason, that it does not know how to do this integral.

3

Looking at Ahmed's Integral -- from Wolfram MathWorld it seems that Mathematica evaluates the generic integrals but not to the same results from the article.

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4

Mathematica did it this way:

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5

Gracias Sorin Suciu.
Ya conocía Ahmed's Integral -- from Wolfram MathWorld.
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
6

How to solve see here and here

Regards M.I

7

Gracias Sorin Suciu.
Ya conocía Ahmed's Integral -- from Wolfram MathWorld.
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