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1 solution
numerator can be written as x^2+1-1=(x^2+1)-1 thus function can be written as ∫1dx-∫(1/(x^2+1))dx within limit 0 to 1.
first function result as x and second as inverse tan x. hence we have
(x-inverse tan x) within limit 0 to 1 which result as 1-π/4=(4-π)/4