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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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    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)$2012-04-22

1 Answers 1

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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 @rhenskyyy2012-04-22