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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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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