 reserve a,x for Real;
 reserve n for Element of NAT;
 reserve A for non empty closed_interval Subset of REAL;
 reserve f,h,f1,f2 for PartFunc of REAL,REAL;
 reserve Z for open Subset of REAL;

theorem
 A c= Z & Z c= dom (ln*(f1+f2)) & f=(id Z)/(a(#)(f1+f2))
 & (for x st x in Z holds h.x=x/a & f1.x=1)
 & a <> 0 & f2=( #Z 2)*h & Z = dom f & f|A is continuous
 implies integral(f,A)=((a/2)(#)(ln*(f1+f2))).(upper_bound A)
                      -((a/2)(#)(ln*(f1+f2))).(lower_bound A)
proof
   assume
A1:A c= Z & Z c= dom (ln*(f1+f2)) & f=(id Z)/(a(#)(f1+f2))
   & (for x st x in Z holds h.x=x/a & f1.x=1)
   & a <> 0 & f2=( #Z 2)*h
   & Z = dom f & f|A is continuous; then
A2:f is_integrable_on A & f|A is bounded by INTEGRA5:10,11;
A3:Z c= dom ((a/2)(#)(ln*(f1+f2))) by A1,VALUED_1:def 5;
A4:for x st x in Z holds f1.x=1 by A1;
A5:for x st x in Z holds h.x=x/a by A1;
then
A6:((a/2)(#)(ln*(f1+f2))) is_differentiable_on Z by A1,A3,A4,SIN_COS9:108;
A7:for x st x in Z holds f.x=x/(a*(1+(x/a)^2))
  proof
  let x;
  assume
A8:x in Z; then
A9:x in dom (f1+f2) by A1,FUNCT_1:11;
   dom (f1+f2) = dom f1 /\ dom f2 by VALUED_1:def 1;then
   dom (f1+f2) c= dom f2 by XBOOLE_1:18;then
A10:x in dom f2 by A9;
  ((id Z)/(a(#)(f1+f2))).x=(id Z).x/(a(#)(f1+f2)).x by A1,A8,RFUNCT_1:def 1
 .=x/(a(#)(f1+f2)).x by A8,FUNCT_1:18
 .=x/(a*(f1+f2).x) by VALUED_1:6
 .=x/(a*(f1.x+f2.x)) by A9,VALUED_1:def 1
 .=x/(a*(1+f2.x)) by A4,A8
 .=x/(a*(1+( #Z 2).(h.x))) by A1,A10,FUNCT_1:12
 .=x/(a*(1+((h.x) #Z 2))) by TAYLOR_1:def 1
 .=x/(a*(1+(h.x)^2)) by FDIFF_7:1
 .=x/(a*(1+(x/a)^2)) by A5,A8;
    hence thesis by A1;
    end;
A11:for x being Element of REAL st x in dom(((a/2)(#)(ln*(f1+f2)))`|Z) holds
  (((a/2)(#)(ln*(f1+f2)))`|Z).x=f.x
   proof
     let x be Element of REAL;
     assume x in dom(((a/2)(#)(ln*(f1+f2)))`|Z);then
A12:  x in Z by A6,FDIFF_1:def 7;then
     (((a/2)(#)(ln*(f1+f2)))`|Z).x
      =x/(a*(1+(x/a)^2)) by A1,A3,A4,A5,SIN_COS9:108
     .=f.x by A7,A12;
     hence thesis;
   end;
   dom(((a/2)(#)(ln*(f1+f2)))`|Z) = dom f by A1,A6,FDIFF_1:def 7;
   then (((a/2)(#)(ln*(f1+f2)))`|Z) = f by A11,PARTFUN1:5;
   hence thesis by A1,A2,A6,INTEGRA5:13;
end;
