Question
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{\text{x}^2-1}\ \text{dx}$
$\int\frac{\text{x}^2+1}{\text{x}^2-1}\ \text{dx}$
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Answer
Let $\text{I}=\int\frac{\text{x}^2+1}{\text{x}^2-1}\ \text{dx}$
$=\int1+\frac{2}{\text{x}^2-1}\ \text{dx}$
$=\int\text{dx}+\int\frac{2\text{dx}}{(\text{x}+1)(\text{x}-1)}$
$=\int\text{dx}+\int\frac{-1}{\text{x}+1}+\frac{1}{\text{x}-1}\ \text{dx}$
$=\text{x}-\log|\text{x}+1|+\log|\text{x}-1|+\text{C}$
$\text{I}=\text{x}+\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
$=\int1+\frac{2}{\text{x}^2-1}\ \text{dx}$
$=\int\text{dx}+\int\frac{2\text{dx}}{(\text{x}+1)(\text{x}-1)}$
$=\int\text{dx}+\int\frac{-1}{\text{x}+1}+\frac{1}{\text{x}-1}\ \text{dx}$
$=\text{x}-\log|\text{x}+1|+\log|\text{x}-1|+\text{C}$
$\text{I}=\text{x}+\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
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