The Formula for the equation is as follows:
$$T(t)=\frac {\int^t(−T_s)ke^{-kt'}dt'+C}{e^{-kt}}$$
This formula is needed to determine the temperature at time $t$, $T(t)$, of an object as it begins to heat. I can't, however, integrate the equation, simply because I cannot make sense of it. Any help would be appreciated so that I cant try and sub my information in and find the answer to the question.
I don't know the source of your equation, but it is given in a rather convoluted form. It may be beneficial to consider where the equation comes from, which should help in clarifying what the equation means. I don't know at what level you understand Newton's law of cooling, so if some of my explanation is pitched too high or too low, I apologize.
Newton's law of cooling/heating is ultimately a differential equation which relates the rate of cooling/heating, i.e. the rate of change of temperature $dT/dt$, with the temperature itself.
Consider a system which is coupled to a heat bath. For example, you might have a cup of coffee as your system, which is interacting with the ambient environment as your heat bath. The temperature of the system is a function of time, say $T(t)$. The underlying assumption is that your heat bath is much larger than your system, and remains at some constant temperature $T_b$.
Physically speaking, we expect the temperature of the system to approach the temperature of the bath as time goes on. We also expect that the rate of cooling/heating is proportional to the temperature difference between the system and bath. This allows us to write down a differential equation $$\frac{dT}{dt} = -k(T-T_b).$$ Here $k$ is some positive constant of proportionality which is related to the physical properties of your system and how easily it heats up and cools down. Note that this equation expresses precisely the conditions we need. For example, if your system temperature is hotter than your heat bath temperature, then $T-T_b > 0$ and $dT/dt < 0$ so your system cools. The hotter your system is, the more rapidly it cools. If they are at the same temperature, $T - T_b =0 $ and your system is in equilibrium with the bath, with $dT/dt = 0$ indicating a non-changing steady-state.
Now it remains to solve this equation, which is a separable first order ODE. We can integrate to get $$\int dT/T-T_b = -k\int dt,$$ which then becomes $$\ln|T-T_b| = -kt + C,$$ for some constant of integration which we will determine later. Taking exponentials, we get $$|T-T_b| = C_0e^{-kt},$$ where $C_0 = e^C$. If we redefine $C_0$ by allowing it to be negative depending on the sign of $T-T_b$, we can write this as $$T(t) = T_b + C_0e^{-kt}.$$ To determine the constant $C_0$, we need an initial condition $T(0) = T_0$, which then tells us $$T(0) = T_0 = T_b + C_0.$$ So our final equation for the temperature is $$T(t) = T_b + (T_0 - T_b)e^{-kt}.$$ If you plot this function, you will see that the temperature $T(t)$ relaxes exponentially to the heat bath temperature $T_b$ as $t\rightarrow \infty$.
Now, for your equation, we have to make a few guesses as to what the terms are. First, the most intuitive guess would be that $T_s$ is some constant temperature. In that case, we would integrate to get $$T(t) = e^{kt}T_s\int (-k)e^{-kt}\ dt + Ce^{kt} = e^{kt}T_se^{-kt} + Ce^{kt} = T_s + Ce^{kt}.$$ This looks very similar to the equation $$T(t) = T_b + C_0e^{-kt},$$ which we had derived earlier, so we should interpret them the same way. In this case, your $k$ is my $-k$, and is the constant of proportionality in the differential equation. Note that $k$ is traditionally and conventionally taken to be positive, so your formula is a little unconventional by that regard (your $k$ must be negative, or the solution is unbounded). Your $T_s$ is my $T_b$, and is the temperature of your heat bath. Your $C$ is my $C_0$, which we've identified to be $$C = C_0 = T_0 - T_s.$$ Of course, this is a complete guess on my part as to what your formula means, but I think it's a reasonable interpretation. In any case, it may be better to stick to the more standard presentation of the law that I've given if possible to avoid any ambiguity.