Question
Evaluate the following integral:
$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
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Answer
$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dx}}{\text{a}^2-\text{b}^2\text{x}}$
$=\frac{1}{\text{b}^2}\times\frac{1}{2\frac{\text{a}}{\text{b}}}\log\Bigg|\frac{\frac{\text{a}}{\text{b}}+\text{x}}{\frac{\text{a}}{\text{b}}-\text{x}}\Bigg|+\text{C}$ $\Big[\therefore\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{x}}{\text{a}-\text{x}}\Big|+\text{C}\Big]$
$=\frac{1}{2\text{ab}}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}\Big|+\text{C}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dx}}{\text{a}^2-\text{b}^2\text{x}}$
$=\frac{1}{\text{b}^2}\times\frac{1}{2\frac{\text{a}}{\text{b}}}\log\Bigg|\frac{\frac{\text{a}}{\text{b}}+\text{x}}{\frac{\text{a}}{\text{b}}-\text{x}}\Bigg|+\text{C}$ $\Big[\therefore\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{x}}{\text{a}-\text{x}}\Big|+\text{C}\Big]$
$=\frac{1}{2\text{ab}}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}\Big|+\text{C}$
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