Concept of $d\left(\frac{1}{T}\right)$ In thermodynamics, I was trying to solve the following integral. 
$$\int_{T_{1}}^{T_{2}} d \ln K=-\frac{\Delta H}{R} \int_{T_{1}}^{T_{2}} d\left(\frac{1}{T}\right)$$
$$\ln K\left(T_{2}\right)-\ln K\left(T_{1}\right)=-\frac{\Delta H}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$$
$$\int_{T_{1}}^{T_{2}} d\left(\frac{1}{T}\right)$$
When I was studying this at school, normally we would have an indefinite integral such as 
$$\int x dx$$
where we have a variable and we are integrating it with respect to a small change to (most of the time) that variable itself, i.e. integrate $x$ with respect to $dx$. 
But in this $\int_{T_{1}}^{T_{2}} d\left(\frac{1}{T}\right)$ I was left rather stuck. 
Are we integrating $1$ with respect to $d\left(\frac{1}{T}\right)$ or is this an incorrect definition of the statement?
I'm just a bit confused by what this means in Layman terms. 
NOTE This is for an introductory chemistry student as well, apologies if the concept seems trivial to others.
 A: The notation may become more intuitive if we look at the Riemann-sum's side. Indeed, much like the ordinary Riemann integral is simply
$$ \int_{a}^{b} f(x) \, \mathrm{d}x \approx \sum_{i=1}^{n} f(x_i) (x_i - x_{i-1}) $$
where the error becomes smaller as the finer partition $\{a = x_0 < x_1 < \cdots < x_n = b\}$ is considered, the Riemann-Stieltjes integral is
$$ \int_{a}^{b} f(x) \, \mathrm{d}g(x) \approx \sum_{i=1}^{n} f(x_i) (g(x_i) - g(x_{i-1})). $$
Here are particular cases:

*

*If $f$ is constant with the value $c$, then all the intermediate terms cancel out, yielding
$$ \int_{a}^{b} c \, \mathrm{d}g(x) \approx \sum_{i=1}^{n} c (g(x_i) - g(x_{i-1})) = c(g(b) - g(a)). $$


*If $g$ is nice (continuously differentiable, for example), then $g(x_i) - g(x_{i-1}) \approx g'(x_i)(x_i - x_{i-1})$ by linear approximation, and so,
$$ \int_{a}^{b} f(x) \, \mathrm{d}g(x) \approx \sum_{i=1}^{n} f(x_i) g'(x_i) (x_i - x_{i-1}) \approx \int_{a}^{b} f(x)g'(x) \, \mathrm{d}x. $$
Of course, the definition of Riemann-Stieltjes integral allows a more general class of increments $\mathrm{d}g(x)$ other than those proportional to $\mathrm{d}x$, which is both theoretically and computationally useful.
A: You're integrating 1 with respect to $\frac1T.$
To make it more simple to understand, let $u=\frac1T.$ Then 
$$\int_{T_1}^{T_2}d(\frac1T)=\int_{\frac{1}{T_1}}^{\frac{1}{T_2}}du=u\Big|_{\frac{1}{T_1}}^{\frac{1}{T_2}}=\frac{1}{T_2}-\frac{1}{T_1}$$
A: note that:
$$\frac{d}{dT}\left[\frac 1T\right]=-\frac{1}{T^2}$$
now if we rearrange we get:
$$d\left[\frac 1T\right]=-\frac{dT}{T^2}$$
applying this to your integral we get:
$$\int_{T_1}^{T_2}d\left[\frac 1T\right]=\int_{T_1}^{T_2}-\frac{dT}{T^2}=\left[\frac 1T\right]_{T_1}^{T_2}=\frac 1{T_2}-\frac{1}{T_1}$$
