1. **Problem statement:** Express the function $$f(x) = \frac{12 + 8x - x^2}{(2 - x)(4 + x^2)}$$ in the form $$\frac{A}{2 - x} + \frac{Bx + C}{4 + x^2}$$ and then show that $$\int_0^1 f(x) \, dx = \ln\left(\frac{25}{2}\right)$$. 2. **Partial fraction decomposition:** We want to find constants $$A$$, $$B$$, and $$C$$ such that: $$ \frac{12 + 8x - x^2}{(2 - x)(4 + x^2)} = \frac{A}{2 - x} + \frac{Bx + C}{4 + x^2} $$ Multiply both sides by the denominator $$ (2 - x)(4 + x^2) $$: $$ 12 + 8x - x^2 = A(4 + x^2) + (Bx + C)(2 - x) $$ 3. **Expand the right side:** $$ A(4 + x^2) + (Bx + C)(2 - x) = 4A + A x^2 + 2Bx + 2C - Bx^2 - Cx $$ Group like terms: $$ = (4A + 2C) + (2B - C)x + (A - B)x^2 $$ 4. **Equate coefficients with the left side:** From $$12 + 8x - x^2$$, coefficients are: - Constant term: 12 - Coefficient of $$x$$: 8 - Coefficient of $$x^2$$: -1 Set up equations: $$ 4A + 2C = 12 \\ 2B - C = 8 \\ A - B = -1 $$ 5. **Solve the system:** From the third equation: $$ A = B - 1 $$ Substitute into the first: $$ 4(B - 1) + 2C = 12 \\ 4B - 4 + 2C = 12 \\ 4B + 2C = 16 $$ From the second equation: $$ 2B - C = 8 \\ C = 2B - 8 $$ Substitute $$C$$ into $$4B + 2C = 16$$: $$ 4B + 2(2B - 8) = 16 \\ 4B + 4B - 16 = 16 \\ 8B = 32 \\ B = 4 $$ Then: $$ C = 2(4) - 8 = 8 - 8 = 0 $$ And: $$ A = 4 - 1 = 3 $$ 6. **Final partial fraction form:** $$ f(x) = \frac{3}{2 - x} + \frac{4x}{4 + x^2} $$ 7. **Integral evaluation:** Compute: $$ \int_0^1 f(x) \, dx = \int_0^1 \frac{3}{2 - x} \, dx + \int_0^1 \frac{4x}{4 + x^2} \, dx $$ 8. **First integral:** Use substitution $$u = 2 - x$$, so $$du = -dx$$: $$ \int_0^1 \frac{3}{2 - x} \, dx = 3 \int_0^1 \frac{1}{2 - x} \, dx = 3 \int_2^1 \frac{-1}{u} \, du = 3 \int_1^2 \frac{1}{u} \, du = 3 [\ln|u|]_1^2 = 3 \ln 2 $$ 9. **Second integral:** Use substitution $$w = 4 + x^2$$, so $$dw = 2x \, dx$$, then: $$ \int_0^1 \frac{4x}{4 + x^2} \, dx = 2 \int_0^1 \frac{2x}{4 + x^2} \, dx = 2 \int_4^5 \frac{1}{w} \, dw = 2 [\ln|w|]_4^5 = 2 \ln \frac{5}{4} $$ 10. **Sum the integrals:** $$ \int_0^1 f(x) \, dx = 3 \ln 2 + 2 \ln \frac{5}{4} = \ln 2^3 + \ln \left(\frac{5}{4}\right)^2 = \ln \left(8 \times \frac{25}{16}\right) = \ln \frac{200}{16} = \ln \frac{25}{2} $$ **Final answer:** $$\boxed{\int_0^1 f(x) \, dx = \ln\left(\frac{25}{2}\right)}$$