In terms of the temperature, the entropy can be defined as $$ \Delta S=\int \frac{dQ}{T}\tag{1} $$ which, as you note, is really a change of entropy and not the entropy itself. Thus, we can write (1) as $$ S(x,T)-S(x,T_0)=\int\frac{dQ(x,T)}{T}\tag{2} $$ But, we are free to set the zero-point of the entropy to anything we want (so as to make it convenient)¹, thus we can use $$S(x,T_0)=0$$ to obtain $$ S(x,T)=\int\frac{dQ(x,T)}{T}\tag{3} $$ If we assume that the heat rise $dQ$ is determined from the heat capacity, $C$, then (3) becomes $$ S(x,T)=\int\frac{C(x,T')}{T'}dT'\tag{4} $$

^{1 This is due to the perfect ordering expected at $T=0$, that is, $S(T=0)=0$, as per the third law of thermodynamics.}

In terms of the temperature, the entropy can be defined as $$ \Delta S=\int \frac{\mathrm dQ}{T}\tag{1} $$ which, as you note, is really a change of entropy and not the entropy itself. Thus, we can write (1) as $$ S(x,T)-S(x,T_0)=\int\frac{\mathrm dQ(x,T)}{T}\tag{2} $$ But, we are free to set the zero-point of the entropy to anything we want (so as to make it convenient)¹, thus we can use $$S(x,T_0)=0$$ to obtain $$ S(x,T)=\int\frac{\mathrm dQ(x,T)}{T}\tag{3} $$ If we assume that the heat rise $\mathrm dQ$ is determined from the heat capacity, $C$, then (3) becomes $$ S(x,T)=\int\frac{C(x,T')}{T'}~\mathrm dT'\tag{4} $$

^{1 This is due to the perfect ordering expected at $T=0$, that is, $S(T=0)=0$, as per the third law of thermodynamics.}