<s>[INST] Which integration is correct? Or are both correct? What's the correct method to integrate the function $\frac{1}{2(3x+1)}$?
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \int\frac{1}{6x+2}dx\\
&= \frac{1}{6}ln(6x+2)+c
\end{align}
$$or$$
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \frac{1}{2}\int\frac{1}{3x+1}dx \\
&= (\frac{1}{2})(\frac{1}{3})ln(3x+1)+c\\
&= \frac{1}{6}ln(3x+1)+c
\end{align}
Both methods seem feasible, however the answers given are different. Is it because the constant c is of a different value?
 [/INST] \begin{align}
\frac{1}{6}\ln(6x+2)+c&= \frac{1}{6}\ln(2(3x+1))+c\\
&=\frac{1}{6}( \ln2 + \ln (3x+1))+c\\
&=\frac{1}{6}( \ln2 + \ln (3x+1))+c\\
&=\frac{1}{6} \ln (3x+1)+c',
\end{align}
with $c'=c+\frac{1}{6}\ln2.$
</s>