Question: Int(ln(1+x)/(1+x^2),x=0..1)

Maple evaluates

Int(ln(1+x)/(1+x^2),x=0..1);

as

1/2*I*dilog(1/2+1/2*I)-1/2*I*dilog(1/2-1/2*I)+1/4*Pi*ln(2)-Catalan

It would be nice if Maple could simplify this to 1/8*Pi*ln(2). If you evalf the integral and use "identify" then Maple does return 1/8*Pi*ln(2).

MMa directly returns 1/8*Pi*ln(2), which seems preferable.

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