let A be closed-interval Subset of REAL ; :: thesis: for f being PartFunc of REAL ,REAL
for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
( f . x = 1 / (x * (1 + ((ln . x) ^2 ))) & ln . x > - 1 & ln . x < 1 ) ) & Z c= dom (arctan * ln ) & Z = dom f & f | A is continuous holds
integral f,A = ((arctan * ln ) . (sup A)) - ((arctan * ln ) . (inf A))
let f be PartFunc of REAL ,REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
( f . x = 1 / (x * (1 + ((ln . x) ^2 ))) & ln . x > - 1 & ln . x < 1 ) ) & Z c= dom (arctan * ln ) & Z = dom f & f | A is continuous holds
integral f,A = ((arctan * ln ) . (sup A)) - ((arctan * ln ) . (inf A))
let Z be open Subset of REAL ; :: thesis: ( A c= Z & ( for x being Real st x in Z holds
( f . x = 1 / (x * (1 + ((ln . x) ^2 ))) & ln . x > - 1 & ln . x < 1 ) ) & Z c= dom (arctan * ln ) & Z = dom f & f | A is continuous implies integral f,A = ((arctan * ln ) . (sup A)) - ((arctan * ln ) . (inf A)) )
assume A1: ( A c= Z & ( for x being Real st x in Z holds
( f . x = 1 / (x * (1 + ((ln . x) ^2 ))) & ln . x > - 1 & ln . x < 1 ) ) & Z c= dom (arctan * ln ) & Z = dom f & f | A is continuous ) ; :: thesis: integral f,A = ((arctan * ln ) . (sup A)) - ((arctan * ln ) . (inf A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: for x being Real st x in Z holds
( ln . x > - 1 & ln . x < 1 ) by A1;
then A4: arctan * ln is_differentiable_on Z by A1, SIN_COS9:117;
A5: for x being Real st x in dom ((arctan * ln ) `| Z) holds
((arctan * ln ) `| Z) . x = f . x

proof

let x be Real; :: thesis: ( x in dom ((arctan * ln ) `| Z) implies ((arctan * ln ) `| Z) . x = f . x )
assume x in dom ((arctan * ln ) `| Z) ; :: thesis: ((arctan * ln ) `| Z) . x = f . x
then A6: x in Z by A4, FDIFF_1:def 8;
then ((arctan * ln ) `| Z) . x = 1 / (x * (1 + ((ln . x) ^2 ))) by A1, A3, SIN_COS9:117
.= f . x by A1, A6 ;
hence ((arctan * ln ) `| Z) . x = f . x ; :: thesis: verum

end;

dom ((arctan * ln ) `| Z) = dom f by A1, A4, FDIFF_1:def 8;
then (arctan * ln ) `| Z = f by A5, PARTFUN1:34;
hence integral f,A = ((arctan * ln ) . (sup A)) - ((arctan * ln ) . (inf A)) by A1, A2, A4, INTEGRA5:13; :: thesis: verum