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?

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.)
