 reserve a,b,x,r for Real;
 reserve y for set;
 reserve n for Element of NAT;
 reserve A for non empty closed_interval Subset of REAL;
 reserve f,g,f1,f2,g1,g2 for PartFunc of REAL,REAL;
 reserve Z for open Subset of REAL;

theorem
 A c= Z & (for x 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).(upper_bound A)
                                          -(arctan*ln).(lower_bound A)
proof
   assume
A1:A c= Z & (for x 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;then
A2:f is_integrable_on A & f|A is bounded by INTEGRA5:10,11;
A3:for x 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 Element of REAL st x in dom ((arctan*ln)`|Z)
holds ((arctan*ln)`|Z).x=f.x
    proof
       let x be Element of REAL;
       assume x in dom ((arctan*ln)`|Z);then
A6:x in Z by A4,FDIFF_1:def 7;then
   ((arctan*ln)`|Z).x=1/(x*(1+(ln.x)^2)) by A1,A3,SIN_COS9:117
                       .=f.x by A1,A6;
   hence thesis;
   end;
  dom ((arctan*ln)`|Z)=dom f by A1,A4,FDIFF_1:def 7;
  then ((arctan*ln)`|Z)= f by A5,PARTFUN1:5;
  hence thesis by A1,A2,A4,INTEGRA5:13;
end;
