<s>[INST] Partial fractions: why does $\int dt \implies t + C$ 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 $A=1/3$
$$\begin{align}
\int \frac{1/3}{y+2} + \frac{1/3}{1-y} \;dy &= \int dt\\
\frac{1}{3}(\ln|y+2| - \ln|1-y|) &= \color{red}{t+C} & \text{?}\\
\ln\left|\frac{y+2}{1-y}\right| &=3(t+C)\\
\frac{y+2}{1-y} = Ce^{3t}
\end{align}$$
 [/INST] 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. 
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