Question
Evaluate the following intregals:
$\int\frac{1}{(\text{x}-1)(\text{x}+1)(\text{x}+2)}\ \text{dx}$

Answer

Let $\text{I}=\int\frac{1}{(\text{x}-1)(\text{x}+1)(\text{x}+2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{\text{x}+2}$
$\Rightarrow1=\text{A}(\text{x}+1)(\text{x}+2)+\text{B}(\text{x}-1)(\text{x}+2)+\text{C}(\text{x}^2-1)$
Put x = 1
$\Rightarrow1=6\text{A}\Rightarrow\text{A}=\frac{1}{6}$
put = -1
$\Rightarrow1=-2\text{B}\Rightarrow\text{B}=-\frac{1}{2}$
put = -2
$\Rightarrow1=3\text{C}\Rightarrow\text{C}=\frac{1}{3}$
So,
$\text{I}=\frac{1}{6}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{3}\int\frac{\text{dx}}{\text{x}+2}$
$\text{I}=\frac{1}{6}\log|\text{x}-1|-\frac{1}{2}\log|\text{x}+1|+\frac{1}{3}\log|\text{x}+2|+\text{C}$

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