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प्रश्न
` ∫ {1}/{a^2 x^2- b^2}dx`
बेरीज
उत्तर
\[\int\frac{dx}{a^2 x^2 - b^2} \]
\[ = \frac{1}{a^2}\int\frac{dx}{x^2 - \left( \frac{b}{a} \right)^2}\]
\[ = \frac{1}{a^2} \times \frac{1}{2\frac{b}{a}} \log \left| \frac{x - \frac{b}{a}}{x + \frac{b}{a}} \right| + C \left[ \therefore \int\frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C \right]\]
` = \text{1}/{2ab} \text{ log }\| \frac{ax - b}{ax + b}| + C `
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