Question

Prove that: \[ \int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log \left( \frac{a + x}{a - x} \right) + C. \]

Hint:

Use the identity: \[ \int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C. \] for integrals of this form.

Answer 1

Step 1: Use Standard Integral Formula
The standard formula for the given integral is: \[ \int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C. \] Step 2: Proof via Partial Fractions
Rewriting: \[ \frac{1}{a^2 - x^2} = \frac{1}{(a - x)(a + x)}. \] Using partial fractions: \[ \frac{1}{(a - x)(a + x)} = \frac{A}{a - x} + \frac{B}{a + x}. \] Solving for \( A \) and \( B \), we obtain: \[ A = \frac{1}{2a}, \quad B = -\frac{1}{2a}. \] Step 3: Integrate
\[ \int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \int \frac{dx}{a + x} - \frac{1}{2a} \int \frac{dx}{a - x}. \] \[ = \frac{1}{2a} \log |a + x| - \frac{1}{2a} \log |a - x|. \] \[ = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C. \]