Which integration is correct? Or are both correct?

Question

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?

Answer 1

\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 $c'=c+\frac{1}{6}\ln2.$

Answer 2

Yes, you are right

Both the feasible answers are correct & the only difference between two is constant of integration