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?
 A: $$\begin{gathered}
  \frac{{x + 2}}
{{{x^3} - x}} = \frac{{x + 2}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}
{x} + \frac{B}
{{x - 1}} + \frac{C}
{{x + 1}} \hfill \\
   = \frac{{A\left( {{x^2} - 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right)}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{\left( {A + B + C} \right){x^2} + \left( {B - C} \right)x - A}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} \hfill \\
   \Rightarrow \left\{ \begin{gathered}
  A + B + C = 0 \hfill \\
  B - C = 1 \hfill \\
   - A = 2 \hfill \\ 
\end{gathered}  \right. \Rightarrow \left\{ \begin{gathered}
  A =  - 2 \hfill \\
  B = 3/2 \hfill \\
  C = 1/2 \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} $$
A: Use partial fractions
$$\frac{x+2}{x^3-x}=\frac{x+2}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}.$$
Solve for $A,B$ and $C$ and then integrate.
A: HINT:
$$\int\frac{x+2}{x^3-x}=\int\frac{-2}{x}+\int\frac{\frac{3}{2}}{x-1}+\int\frac{\frac{1}{2}}{x+1}$$
Can u do it from here?
A: The Heaviside cover-up method for solving partial fraction decompositions deserves to be more widely known. We want to find $A,B,C$ in this equation:
$$\frac{x+2}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$$
To find $A$, multiply by $x$:
$$\frac{x+2}{(x-1)(x+1)}=A+\frac{Bx}{x-1}+\frac{Cx}{x+1}$$
Now put $x=0$ to get $\dfrac{2}{-1}=A$.  
It looks like a swindle, because the starting equation is not valid when $x=0$. But if this bothers you, you can make it rigorous by taking the limit as $x \to 0$ in the second equation.
To find $B$, multiply by $x-1$ and put $x=1$ to get $\dfrac{3}{2}=B$.
To find $C$, multiply by $x+1$ and put $x=-1$ to get $\dfrac{1}{2}=C$.
(Things are not quite so simple if the denominator has a repeated root, but it's still doable. See the link for details.)
