1. **State the problem:** Evaluate the integral $$\int \frac{5x^2}{x^4 - 1} \, dx$$. 2. **Identify the integral type:** This is a rational function integral where the denominator can be factored. 3. **Factor the denominator:** $$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x-1)(x+1)(x^2 + 1)$$ 4. **Set up partial fraction decomposition:** $$\frac{5x^2}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx + D}{x^2 + 1}$$ 5. **Multiply both sides by the denominator:** $$5x^2 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx + D)(x^2 - 1)$$ 6. **Expand and collect like terms:** $$5x^2 = A(x^3 + x^2 + x + 1) + B(x^3 - x^2 + x - 1) + (Cx + D)(x^2 - 1)$$ 7. **Expand further:** $$5x^2 = A x^3 + A x^2 + A x + A + B x^3 - B x^2 + B x - B + C x^3 - C x + D x^2 - D$$ 8. **Group terms by powers of x:** $$5x^2 = (A + B + C) x^3 + (A - B + D) x^2 + (A + B - C) x + (A - B - D)$$ 9. **Set up system of equations by matching coefficients:** - Coefficient of $x^3$: $A + B + C = 0$ - Coefficient of $x^2$: $A - B + D = 5$ - Coefficient of $x$: $A + B - C = 0$ - Constant term: $A - B - D = 0$ 10. **Solve the system:** From equations for $x^3$ and $x$: $$A + B + C = 0$$ $$A + B - C = 0$$ Add both: $2A + 2B = 0 \Rightarrow A + B = 0 \Rightarrow B = -A$ Substitute $B = -A$ into $A + B + C = 0$: $$A - A + C = 0 \Rightarrow C = 0$$ From constant term and $x^2$ term: $$A - B - D = 0$$ $$A - (-A) - D = 0 \Rightarrow 2A - D = 0 \Rightarrow D = 2A$$ $$A - B + D = 5$$ $$A - (-A) + 2A = 5 \Rightarrow 4A = 5 \Rightarrow A = \frac{5}{4}$$ Then: $$B = -\frac{5}{4}, C = 0, D = 2 \times \frac{5}{4} = \frac{5}{2}$$ 11. **Rewrite the integral with partial fractions:** $$\int \frac{5x^2}{x^4 - 1} dx = \int \frac{5/4}{x-1} dx + \int \frac{-5/4}{x+1} dx + \int \frac{(0)x + 5/2}{x^2 + 1} dx$$ 12. **Integrate each term:** $$\int \frac{5/4}{x-1} dx = \frac{5}{4} \ln|x-1| + C_1$$ $$\int \frac{-5/4}{x+1} dx = -\frac{5}{4} \ln|x+1| + C_2$$ $$\int \frac{5/2}{x^2 + 1} dx = \frac{5}{2} \arctan(x) + C_3$$ 13. **Combine all results:** $$\int \frac{5x^2}{x^4 - 1} dx = \frac{5}{4} \ln|x-1| - \frac{5}{4} \ln|x+1| + \frac{5}{2} \arctan(x) + C$$ **Final answer:** $$\boxed{\frac{5}{4} \ln|x-1| - \frac{5}{4} \ln|x+1| + \frac{5}{2} \arctan(x) + C}$$