integration of fractions i am trying to integrate following equation
$$ \int\frac 1{(x^2-1)\cdot (x+2)}\,dx$$
i can represent $(x^2-1)=(x-1)(x+1)$  so,it would be converted in the following form
$$\int\frac1{(x^2-1)(x+2)}\,dx=\int \frac1{(x-1)(x+1)(x+2)}\,dx$$
or it is equal  $$\int \frac1{(x-1)(x^2+3x+2)}\,dx$$  last one  we can decompose  into form
$$ \frac1{(x-1)(x^2+3x+2)}=\frac A{x-1}+\frac{Cx+D}{x^2+3x+2}$$  am i right?or did i miss some term?

I think you can decompose it like this:
$$
\frac{1}{(x^2-1)\cdot(x+2)}=\frac{a}{x-1}+\frac{b}{x+1}+\frac{c}{x+2}
$$
Thus we can solve the following equations:
$$
a+b+c=0\\3a+b=0\\2a-2b-c=1
$$
getting $a=1/6,b=-1/2,c=1/3$.
Therefore,
$$
\int\frac{dx}{(x^2-1)\cdot(x+2)}\\=\int\frac{1}{6}\cdot\frac{dx}{x-1}-\int\frac{1}{2}\cdot\frac{dx}{x+1}+\int\frac{1}{3}\cdot\frac{dx}{x+2}\\=\frac{1}{6}\cdot \log(x-1)-\frac{1}{2}\cdot \log(x+1)+\frac{1}{3}\cdot \log(x+2).
$$
