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?

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Why don't you use $\frac 1{(x-1)(x+1)(x+2)}$ and decompose as $\frac \alpha{x-1} + \frac\beta{x+1} + \frac\gamma{x+2}$? – martini Apr 22 '12 at 8:28
i was thinking that,because we have quadratic term,i would lose some fraction – dato datuashvili Apr 22 '12 at 8:35
In partial fractions, you typically don't use quadratics in the denominator unless they can't be further reduced. – Mike Apr 22 '12 at 8:53
thanks very much @Mike – dato datuashvili Apr 22 '12 at 8:56
The common name for this type of integration is Integration using Partial Fractions. This is a special case of integrating fractions of the general form of $f(x)/g(x)$ – NoChance Apr 22 '12 at 9:17

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,

$$\int\frac{dx}{(x^2-1)\cdot(x+2)}\\=\int\frac{1}{6}\cdot\frac{dx}{x-1}-\int\frac{1}{2}\cdot\frac{dx}{x+1}+\int\frac{1}{3}\cdot\frac{dx}{x+2}\\=\frac{1}{6}\cdot \log(x-1)-\frac{1}{2}\cdot \log(x+1)+\frac{1}{3}\cdot \log(x+2).$$

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so it means that,my multiplication of (x+1) by (x+2) was not necessary,thanks a lot of @rhenskyyy – dato datuashvili Apr 22 '12 at 8:51