Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did:
$$\int \frac{x+2}{x^3-x} dx$$
then:
$$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$  
$$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$
$$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$
now I need to resolve one integral at the time so:
$$\int \frac{1}{x^2-1}$$ with x = t I have:
$$\int \frac{1}{t^2-1}$$ 
Now I have no idea about how to procede with this...any help?

$$\begin{gathered}
  \frac{{x + 2}}
{{{x^3} - x}} = \frac{{x + 2}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}
{x} + \frac{B}
{{x - 1}} + \frac{C}
{{x + 1}} \hfill \\
   = \frac{{A\left( {{x^2} - 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right)}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{\left( {A + B + C} \right){x^2} + \left( {B - C} \right)x - A}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} \hfill \\
   \Rightarrow \left\{ \begin{gathered}
  A + B + C = 0 \hfill \\
  B - C = 1 \hfill \\
   - A = 2 \hfill \\ 
\end{gathered}  \right. \Rightarrow \left\{ \begin{gathered}
  A =  - 2 \hfill \\
  B = 3/2 \hfill \\
  C = 1/2 \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} $$
