Partial fractions: why does $\int dt \implies t + C$

Question

I am working on a partial fraction problem here, I understand everything in the problem except $t+C$, so I'd like to know where did the $t+C$ come from ?

I want to solve this integral $$ \int \frac{dy}{(y+2)(1-y)} = \int dt $$

$$\begin{align} 1 &= \frac{A}{y+2} + \frac{B}{1-y} \\ 1 &= A(1-y) + B(y+2) \end{align}$$

Let $y=1$, then $1=B(3)$ and $B=1/3$.

Let $y=-2$, then $1=A(3)$ and $A=1/3$

$$\begin{align} \int \frac{1/3}{y+2} + \frac{1/3}{1-y} \;dy &= \int dt\\ \frac{1}{3}(\ln|y+2| - \ln|1-y|) &= \color{red}{t+C} & \text{?}\\ \ln\left|\frac{y+2}{1-y}\right| &=3(t+C)\\ \frac{y+2}{1-y} = Ce^{3t} \end{align}$$

Answer 1

Your original equation is $$\int \frac{dy}{(y+2)(1-y)} = \int \,dt$$ $t + C$ comes from integrating the righthand side of the original equation: $$\int \,dt = t + C$$

The $C$ is the constant of integration.

Answer 2

Think of $\int dt$ as $\int1dt$. The integral of $1$ with respect to any variable is that variable, so in this case, $\int1dt$ is $t+C$. Where $C$ is a constant.