Question

The value of \[ \int \frac{dx}{x^2 - a^2} \] will be:

• A. $\frac{1}{2a^2} \log \left| \frac{x - a}{x + a} \right| + C$
• B. $\frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C$
• C. $\frac{1}{4a} \log \left| \frac{x + a}{x - a} \right| + C$
• D. $\frac{1}{4a^2} \log \left| \frac{x + a}{x - a} \right| + C$

Hint:

When dealing with integrals of the form \(\frac{1}{x^2 - a^2}\), use partial fraction decomposition and apply standard integral formulas.

Answer 1

The Correct Option is B

We are asked to evaluate the integral: \[ \int \frac{dx}{x^2 - a^2}. \]

Step 1: Factor the denominator.
We can factor the denominator using the difference of squares formula: \[ x^2 - a^2 = (x - a)(x + a). \] Thus, the integral becomes: \[ \int \frac{dx}{(x - a)(x + a)}. \]

Step 2: Use partial fraction decomposition.
We can decompose the integrand as: \[ \frac{1}{(x - a)(x + a)} = \frac{A}{x - a} + \frac{B}{x + a}. \] Multiplying both sides by \((x - a)(x + a)\), we get: \[ 1 = A(x + a) + B(x - a). \] Expanding and simplifying: \[ 1 = A(x) + Aa + B(x) - Ba, \] \[ 1 = (A + B)x + (Aa - Ba). \] Equating the coefficients of \(x\) and the constants, we get: \[ A + B = 0, Aa - Ba = 1. \] From \(A + B = 0\), we have \(B = -A\). Substituting into the second equation: \[ Aa - A(-a) = 1, \] \[ 2aA = 1 $\Rightarrow$ A = \frac{1}{2a}, B = -\frac{1}{2a}. \]

Step 3: Substitute back into the integral.
Now the integral becomes: \[ \int \left( \frac{1}{2a} \left( \frac{1}{x - a} \right) - \frac{1}{2a} \left( \frac{1}{x + a} \right) \right) dx. \] This simplifies to: \[ \frac{1}{2a} \int \left( \frac{1}{x - a} - \frac{1}{x + a} \right) dx. \]

Step 4: Integrate each term.
We know that the integral of \(\frac{1}{x - a}\) is \(\log|x - a|\) and the integral of \(\frac{1}{x + a}\) is \(\log|x + a|\). Thus, we have: \[ \frac{1}{2a} \left( \log|x - a| - \log|x + a| \right). \] This simplifies to: \[ \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C. \]

Step 5: Conclusion.
The correct answer is (B) \(\frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C\).