Minor mistake computing $\int \frac{1}{x^3+2x^2-3x} \; dx$? I'm trying to compute:
$$\int \frac{1}{x^3+2x^2-3x} \; dx$$
Until now, I did the following: Factoring:
$$x^3+2x^2-3x=x(x-1)(x+3)$$
To obtain the parcial fractions:
$$\frac{a}{x}+\frac{b}{x+3}+\frac{c}{x-1}=\frac{1}{x(x-1)(x+3)}$$
$$a(x-1)(x+3)+bx(x-1)+cx(x+3)=1$$
$$-3 a + 2 a x - b x + 3 c x + a x^2 + b x^2 + c x^2 = 1$$
This give me the system:
$$\begin{eqnarray*}
  {a+b+c}&=&{0} \\ 
  {2a-b-3c}&=&{0} \\ 
  {-3a}&=&{1} 
\end{eqnarray*}$$
Solving it yields: 
$$a= -\cfrac{1}{3}\quad b=\cfrac{5}{6} \quad c=-\cfrac{1}{3}$$
And then:
$$\int \frac{-1/3}{x}+\frac{5/6}{x+3}+\frac{-1/2}{x-1} \; dx$$
$$\int \frac{-1/3}{x} \; dx+ \int \frac{5/6}{x+3}dx+ \int \frac{-1/2}{x-1} \; dx + C$$
$$\frac{-1}{3}\int \frac{1}{x} \; dx+ \frac{5}{6} \int \frac{1}{x+3}dx+ \frac{-1}{2} \int \frac{1}{x-1} \; dx + C$$
Which should be:
$$\frac{-1}{3} \log(x)+ \frac{5}{6} \log(x+3)+ \frac{-1}{2} \log(x-1)+ C$$
But Mathematics gives me: 
$$\frac{1}{4} \log (1-x)-\frac{\log (x)}{3}+\frac{1}{12} \log (x+3)$$
But I have no idea of what I did wrong.

It should be $2a-b+3c=0$, not $2a-b-3c=0$.
As a side note, if you want to solve $a(x-1)(x+3)+bx(x-1)+cx(x+3)=1$, then it's a lot easier to plug in $x=0, 1, -3$ instead of expanding.
