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

theorem
  A c= Z & f = ln*f1 & (for x st x in Z holds f1.x=x+a & f1.x>0) & dom (
id Z -(2*a)(#)f) = Z & Z = dom f2 & (for x st x in Z holds f2.x = (x-a)/(x+a))
& f2|A is continuous implies integral(f2,A) = (id Z -(2*a)(#)f).(upper_bound A)
  -(id Z -(2*a)(#)f).(lower_bound A)
proof
  assume that
A1: A c= Z and
A2: ( f = ln*f1 & for x st x in Z holds f1.x=x+a & f1.x>0 )& dom (id Z -
  (2*a) (#)f) = Z and
A3: Z = dom f2 and
A4: for x st x in Z holds f2.x = (x-a)/(x+a) and
A5: f2|A is continuous;
A6: f2 is_integrable_on A by A1,A3,A5,INTEGRA5:11;
A7: (id Z -(2*a)(#)f) is_differentiable_on Z by A2,FDIFF_4:6;
A8: for x being Element of REAL
st x in dom ((id Z -(2*a)(#)f)`|Z) holds ((id Z -(2*a)(#)f)`|Z).x =
  f2.x
  proof
    let x be Element of REAL;
    assume x in dom ((id Z -(2*a)(#)f)`|Z);
    then
A9: x in Z by A7,FDIFF_1:def 7;
    then ((id Z -(2*a)(#)f)`|Z).x = (x-a)/(x+a) by A2,FDIFF_4:6
      .= f2.x by A4,A9;
    hence thesis;
  end;
  dom ((id Z -(2*a)(#)f)`|Z) = dom f2 by A3,A7,FDIFF_1:def 7;
  then ((id Z -(2*a)(#)f)`|Z) = f2 by A8,PARTFUN1:5;
  hence thesis by A1,A3,A5,A6,A7,INTEGRA5:10,13;
end;
