Order of factors in partial decomposition Is there a protocol for deciding which denominator fraction goes under A and which goes under B during partial decomposition?
Doing this question: integral $(5x-5)/(3x^2-8x-3)$ I factored the denominator and got $(3x+1)(x-3)$, so I moved on with $A/(3x+1) + B/(x-3)$.
This came out as $A=2, B=1$, so my final answer answer was $2\ln|x-3| + 1/3\ln|3x+1|$.
However the solution in the textbook had it reversed, with $A=1, B=2$, and the final answer was therefore $\ln|x-3| + 2/3 \ln|3x+1|$.
Are the two answers the same? And thus just need some natural log algebra to reflect this? Or is there, as asked, a definite way of saying which factor goes under A and which B?
 A: In the way you have written the partial fractions, $A=1, B=2$ doesn't work. Check the numerator: for instance, you get $7x$ instead of $5x$.
As pointed out in one of the comments, the textbook has probably swapped the factors, which explains why you get swapped values.
A: As JG123's comment stated, you just flipped the $A$ and $B$ in your calculations. Note you have
$$\begin{equation}\begin{aligned}
\frac{5x-5}{3x^2 -8x -3} & = \frac{A}{3x+1} + \frac{B}{x-3} \\
\frac{5x-5}{3x^2 -8x -3} & = \frac{A(x-3) + B(3x + 1)}{(3x + 1)(x-3)} \\
5x-5 & = A(x-3) + B(3x + 1)
\end{aligned}\end{equation}\tag{1}\label{eq1}$$
This leads to
$$5x = (A + 3B)x \implies 5 = A + 3B \tag{2}\label{eq2}$$
$$-5 = -3A + B \tag{3}\label{eq3}$$
Multiplying \eqref{eq2} by $3$ and adding \eqref{eq3} gives
$$15 - 5 = 9B + B \implies B = 1 \tag{4}\label{eq4}$$
Using this in \eqref{eq3} gives $A = 2$.
Thus, using this in the integration, the final answer will then match what the book gave of $\ln|x-3| + 2/3 \ln|3x+1|$.
As for which partial fraction to use for $A$ and which for $B$, this is arbitrary as both $A$ and $B$ are dummy variables. The book just seems to have used them in the opposite order to what you did.
