Question

Find: \[ \int x^2 \log(x^2 - 1) \, dx. \]

Hint:

For integrals involving logarithms and polynomials, use integration by parts. Carefully simplify the integrals and always check if partial fractions can help in reducing the complexity.

Answer 1

Given integral: \[ \int x^2 \log(x^2 - 1) \, dx \] Using integration by parts: \[ \int u \, dv = uv - \int v \, du. \] Let: \[ u = \log(x^2 - 1) \quad \text{and} \quad dv = x^2 \, dx. \] Then, we have: \[ du = \frac{2x}{x^2 - 1} \, dx \quad \text{and} \quad v = \frac{x^3}{3}. \] Substitute these into the integration by parts formula: \[ \int x^2 \log(x^2 - 1) \, dx = \frac{x^3}{3} \log(x^2 - 1) - \int \frac{x^3}{3} \cdot \frac{2x}{x^2 - 1} \, dx. \] Simplifying the second term: \[ = \frac{x^3}{3} \log(x^2 - 1) - \frac{2}{3} \int \frac{x^4 - 1}{x^2 - 1} \, dx. \] Now, split the integral: \[ = \frac{x^3}{3} \log(x^2 - 1) - \frac{2}{3} \int \left( x^2 + 1 - \frac{1}{x^2 - 1} \right) \, dx. \] Integrating each term: \[ = \frac{x^3}{3} \log(x^2 - 1) - \frac{2}{3} \left( \int (x^2 + 1) \, dx - \int \frac{1}{x^2 - 1} \, dx \right). \] \[ = \frac{x^3}{3} \log(x^2 - 1) - \frac{2}{3} \left( \frac{x^3}{3} + \frac{x}{2} \log |x - 1| + \frac{x}{2} \log |x + 1| \right) + C. \] Conclusion: The solution to the integral is: \[ \frac{x^3}{3} \log(x^2 - 1) - \frac{2}{3} \left( \frac{x^3}{3} + \frac{x}{2} \log |x - 1| + \frac{x}{2} \log |x + 1| \right) + C. \]