integral parameterization in MATLAB [duplicate] - matlab

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Using MATLAB integral with anonymous functions
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I want to use integral function which is parameterized. but it gives an error. the upper limit in integral function is a parameter.
my code is here:
f1=#(t) (25.46/t)+0.01519
antf1=integral(f1,298.15,t)
but, for example, the following code does not give any error:
f2=#(t) 25.46*(1.519*t/100)
antf2=integral(f2,298.15,t)
thank you all.
UPDATE 1
the complete code is here:
clc
clear
i=1;
j=1;
a=1.4979e-06;
b=1.9829e-05;
r=0.00008314;
for t=173.15:10:423.15
tr=t/154.6;
for p=1:1:100
pr=p/50.45;
aa=( 1.4084*(1-(t/154.6)^0.5 ))^2;
eq=[p -((3*b*p)+(r*t)) -((r*2*t*b)-(aa*a*b)+(3*p*(b^2))) (((b^3)*p)-(aa*a*b)+(r*t*(b^2)))];
rr=roots(eq);
rq=rr(1);
v(i,j)=rq;
bb=(r*154.6/50.45)*((0.083-(0.422/(tr^1.6)))+(0.022*(0.139-(0.172/(tr^4.2)))));
zv(i,j)=1+((pr/tr)*(bb*50.45/(r*154.6)));
%z(i,j)=(v(i,j)*p)/(r*t);
fun1= #(t) 25.46+(1.519*t/100) ;%-(0.7151*(t^2)/100000)+(1.311*(t^3)/1000000000);
fun2= #(p) ((p/(t*r))*(r*154.6/50.45)*((0.6752*(154.6^1.6)/(t^0.6))+((0.7224*(154.6^4.2)/(t^3.2)))))-(p*bb/(r*(t^2)));
infun1=integral(fun1,298.15,t);
infun2=integral(fun2,1,p);
h(i,j)= infun1 - (r*(t^2))*infun2;
f1= #(t) (25.46/t)+0.01519 ;
f2= #(p) ((1/(t*r))*(r*154.6/50.45)*((0.6752*(154.6^1.6)/(t^0.6))+((0.7224*(154.6^4.2)/(t^3.2)))))-(1*bb/(r*(t^2)));
f3= #(p) (p*t/(50.45*154.6))*(bb*50.45/(r*154.6));
anf3=integral(f1,298.15,t); %%%%%%%%%%% LINE 28 %%%%%%%%%%%
anf4=integral(f2,1,p);
anf5=integral(f3,1,p);
s(i,j)=anf3-(r*log(p/1))+(r*((-1)*t*anf4)-anf5);
i=i+1;
end
i=1;
j=j+1;
end
v;
%z;
zv;
h;
s;
the error message is shown in the following:
Error using /
Matrix dimensions must agree.
Error in code>#(t)(25.46/t)+0.01519
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 75)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in integral (line 88)
Q = integralCalc(fun,a,b,opstruct);
Error in code (line 28)
anf3=integral(f1,298.15,t);