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

Answer

Let $\text{I}=\int\frac{\text{x}^2+1}{\text{x}(\text{x}^2-1)}\ \text{dx}=\int\frac{\text{x}^2+1}{\text{x}(\text{x}+1)(\text{x}-1)}\ \text{dx}$
Let $\frac{\text{x}^2+1}{\text{x}(\text{x}+1)(\text{x}-1)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{\text{x}-1}$
$\Rightarrow\text{x}^2+1=\text{A}(\text{x}+1)(\text{x}-1)+\text{Bx}(\text{x}-1)+\text{Cx}(\text{x}+1)$
Put x = 0
⇒ 1 = -A ⇒ A = -1
Put x = -1
⇒ 2 = 2B ⇒ B = 1
Put x = 1
⇒ 2 = 2C ⇒ C = 1
Thus,
$\text{I}=-\int\frac{\text{dx}}{\text{x}}+\int\frac{\text{dx}}{\text{x}+1}+\int\frac{\text{dx}}{\text{x}-1}$
$=-\log|\text{x}|+\log|\text{x}+1|+\log|\text{x}-1|+\text{C}$
$\text{I}=\log\Big|\frac{\text{x}^2-1}{\text{x}}\Big|+\text{C}$

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