Question

Evaluate \( \int \frac{dx}{x(x^2-1)} \).

Hint:

For rational functions, decompose into partial fractions for simpler integration.

Answer 1

Solve \( \int \frac{dx}{x(x^2 - 1)} \).

Step 1: Partial Fraction Decomposition
Factor the denominator: \[ \frac{1}{x(x^2 - 1)} = \frac{1}{x(x - 1)(x + 1)} \] Set up the decomposition: \[ \frac{1}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1} \] Clear the denominators: \[ 1 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1) \]
Step 2: Determine coefficients
Equate coefficients of like powers of \( x \): \[ A + B + C = 0, \quad B - C = 0, \quad -A = 1 \] The coefficients are: \[ A = -1, \quad B = \frac{1}{2}, \quad C = \frac{1}{2} \]
Step 3: Integrate terms
The integral becomes: \[ \int \left( \frac{-1}{x} + \frac{1/2}{x - 1} + \frac{1/2}{x + 1} \right) dx \] Performing the integration: \[ = -\ln|x| + \frac{1}{2} \ln|x - 1| + \frac{1}{2} \ln|x + 1| + C \]
Step 4: Combine logarithmic terms
The final result is: \[ \int \frac{dx}{x(x^2 - 1)} = \frac{1}{2} \ln \left| \frac{x^2 - 1}{x^2} \right| + C \]