Question

Find the value of \(\int \frac{1}{x^2 - a^2} dx\).

Hint:

This is one of the fundamental integration formulas and is worth memorizing to save time.
\(\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C\)
Also remember the related formula:
\(\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \ln\left|\frac{a+x}{a-x}\right| + C\)

Answer 1

Step 1: Understanding the Concept:
This is a standard indefinite integral of a rational function. The standard method to solve it from first principles is using partial fraction decomposition. Alternatively, one can state the well-known formula directly. We will show the derivation using partial fractions.
Step 2: Key Formula or Approach:
1. Factor the denominator: \(x^2 - a^2 = (x - a)(x + a)\).
2. Decompose the fraction \(\frac{1}{(x-a)(x+a)}\) into partial fractions: \(\frac{A}{x-a} + \frac{B}{x+a}\).
3. Solve for the constants A and B.
4. Integrate the resulting simpler fractions.
Step 3: Detailed Explanation or Calculation:
1. Partial Fraction Decomposition: \[ \frac{1}{x^2 - a^2} = \frac{1}{(x-a)(x+a)} = \frac{A}{x-a} + \frac{B}{x+a} \] To find A and B, we multiply by the common denominator: \[ 1 = A(x+a) + B(x-a) \] We can find the coefficients by substituting convenient values for \(x\).
Let \(x = a\): \[ 1 = A(a+a) + B(a-a) \implies 1 = 2aA \implies A = \frac{1}{2a} \] Let \(x = -a\): \[ 1 = A(-a+a) + B(-a-a) \implies 1 = -2aB \implies B = -\frac{1}{2a} \] So, the decomposition is: \[ \frac{1}{x^2 - a^2} = \frac{1}{2a(x-a)} - \frac{1}{2a(x+a)} \] 2. Integration: Now we integrate the expression: \[ \int \frac{1}{x^2 - a^2} dx = \int \left( \frac{1}{2a(x-a)} - \frac{1}{2a(x+a)} \right) dx \] \[ = \frac{1}{2a} \int \left( \frac{1}{x-a} - \frac{1}{x+a} \right) dx \] Using the standard integral \(\int \frac{1}{u} du = \ln|u|\): \[ = \frac{1}{2a} (\ln|x-a| - \ln|x+a|) + C \] Using the logarithm property \(\ln M - \ln N = \ln(M/N)\): \[ = \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C \] Step 4: Final Answer:
The value of the integral is \(\frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C\).