Simple Partial Fractions Question For practice, I am integrating,
$$\int \frac{x}{3x^2 + 8x -3}dx$$
So, I can then factor it as,
$$\int \frac{x}{(3x-1)(x+3)}dx$$
By partial fractions, I decompose
$$\frac{x}{(3x-1)(x+3)}= \frac{A}{3x-1} + \frac{B}{x+3}$$
For finding $A$, I multiply both sides by $3x-1$, which gives
$$\frac{x(3x-1)}{(3x-1)(x+3)} = \frac{A(3x-1)}{3x-1} + \frac{B(3x-1)}{x+3}$$
So, we have that
$$\frac{x}{(x+3)} = A + \frac{B(3x-1)}{x+3}$$
Letting $3x-1=0$, we have that $x=\frac{1}{3}$, so then
$$\frac{\frac{1}{3}}{(\frac{1}{3}+3)} = A$$
Thus, we have that $A=\frac{1}{10}$. For determining $B$, we then multiply both sides by $x+3$ and receive, as a similar process to the previous,
$$\frac{x(x+3)}{(3x-1)(x+3)} = \frac{A(x+3)}{3x-1} + \frac{B(x+3)}{x+3}$$
Then,
$$\frac{x}{3x-1} = \frac{A(x+3)}{3x-1} + B$$
So, if we let $x+3=0$, we then have that $x=-3$ and so,
$$\frac{-3}{3(-3)-1}=B$$
So, we then have that $B=\frac{3}{10}$. Thus, our original integral can then be written as,
$$\int \frac{x}{(3x-1)(x+3)}dx = \int \frac{1}{10(3x-1)} + \frac{3}{10(x+3)} dx$$
We can, by splitting up the integral find,
$$\int \frac{x}{(3x-1)(x+3)}dx = \frac{1}{10} \int \frac{1}{3x-1} dx + \frac{3}{10} \int \frac{1}{x+3} dx$$
Thus, we conclude that,
$$\int \frac{x}{3x^2 + 8x -3}dx = \frac{\ln|3x-1|}{30} + \frac{3 \ln|x+3|}{10} + C$$
Wolframalpha shows that, the answer is:
$$\frac{1}{30}(\ln(1-3x)+ 9 \ln(3+x)) +C$$
What am I doing wrong, did I miss a negative sign somewhere?

Wolfram forgot the absolute value signs. Further both answers are the same.
$$\frac{9}{30}=\frac{3}{10}$$
