To evaluate integrals of the form \(\int \frac{1}{x^2 - a^2} \, dx\), a standard approach is to use trigonometric substitution. Let \(x = a\sec\theta\), which implies \(dx = a\sec\theta\tan\theta \, d\theta\), and Since \(x = a\sec\theta\), we have \(x^2 = a^2\sec^2\theta\). Therefore, \(x^2 - a^2 = a^2\sec^2\theta - a^2 = a^2(\sec^2\theta - 1) = a^2\tan^2\theta\)
Substituting into the integral: \( \int \frac{dx}{x^2 - a^2}\) \(\int \frac{dx}{x^2 - a^2} = \int \frac{a\sec\theta \tan\theta , d\theta}{a^2 \tan^2\theta} = \frac{1}{a} \int \frac{\sec\theta}{\tan\theta} , d\theta \)
Now using the identity: \( \frac{\sec\theta}{\tan\theta} = \cosec\theta \quad\) (since \( \sec\theta = \frac{1}{\cos\theta}, \tan\theta = \frac{\sin\theta}{\cos\theta},\) so \(\frac{\sec\theta}{\tan\theta} = \frac{1/\cos\theta}{\sin\theta/\cos\theta} = \frac{1}{\sin\theta} = \cosec\theta \) Therefore: \( \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{a} \int \cosec\theta \, d\theta \)
And we know: \( \int \cosec\theta , d\theta = \ln \left| \cosec\theta + \cot\theta \right| + C \)
Now return to the original variable \(x\) using \(x = a\sec\theta\), which gives \(\sec\theta = \frac{x}{a}\). From this: \( \cosec\theta = \frac{\sqrt{x^2 - a^2}}{x}, \quad \cot\theta = \frac{a}{\sqrt{x^2 - a^2}} \)
Substituting these back into the logarithmic expression and simplifying yields: \( \int \frac{1}{x^2 - a^2} , dx = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C \)
Therefore:
\( \boxed{\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + c} \)
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