1. **Problem statement:** Find the volume of the solid formed by rotating the region under the curve $$y=\frac{5}{x^2+5x+6}$$ from $$x=0$$ to $$x=1$$ about the x-axis. 2. **Formula for volume about the x-axis:** $$V=\pi \int_a^b [f(x)]^2 dx$$ This formula calculates the volume of a solid of revolution by integrating the square of the function times $$\pi$$ over the interval. 3. **Rewrite the function:** Factor the denominator: $$x^2+5x+6=(x+2)(x+3)$$ So, $$y=\frac{5}{(x+2)(x+3)}$$ 4. **Set up the integral:** $$V=\pi \int_0^1 \left(\frac{5}{(x+2)(x+3)}\right)^2 dx=25\pi \int_0^1 \frac{1}{(x+2)^2 (x+3)^2} dx$$ 5. **Partial fraction decomposition:** We want to express $$\frac{1}{(x+2)^2 (x+3)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{x+3} + \frac{D}{(x+3)^2}$$ Multiply both sides by $$ (x+2)^2 (x+3)^2 $$: $$1 = A(x+2)(x+3)^2 + B(x+3)^2 + C(x+3)(x+2)^2 + D(x+2)^2$$ 6. **Find coefficients by substituting values:** - Let $$x=-2$$: $$1 = B(1)^2 = B \Rightarrow B=1$$ - Let $$x=-3$$: $$1 = D( -1)^2 = D \Rightarrow D=1$$ 7. **Expand and equate coefficients:** Expand terms and equate coefficients of powers of $$x$$ to solve for $$A$$ and $$C$$: After simplification, we find: $$A = -1, \quad C = -1$$ 8. **Rewrite the integral:** $$\int_0^1 \frac{1}{(x+2)^2 (x+3)^2} dx = \int_0^1 \left( \frac{-1}{x+2} + \frac{1}{(x+2)^2} - \frac{1}{x+3} + \frac{1}{(x+3)^2} \right) dx$$ 9. **Integrate term by term:** - $$\int \frac{1}{(x+a)^2} dx = -\frac{1}{x+a} + C$$ - $$\int \frac{1}{x+a} dx = \ln|x+a| + C$$ So, $$\int_0^1 \frac{-1}{x+2} dx = -[\ln(x+2)]_0^1 = -\ln 3 + \ln 2$$ $$\int_0^1 \frac{1}{(x+2)^2} dx = \left[-\frac{1}{x+2}\right]_0^1 = -\frac{1}{3} + \frac{1}{2} = \frac{1}{6}$$ $$\int_0^1 \frac{-1}{x+3} dx = -[\ln(x+3)]_0^1 = -\ln 4 + \ln 3$$ $$\int_0^1 \frac{1}{(x+3)^2} dx = \left[-\frac{1}{x+3}\right]_0^1 = -\frac{1}{4} + \frac{1}{3} = \frac{1}{12}$$ 10. **Sum all integrals:** $$I = (-\ln 3 + \ln 2) + \frac{1}{6} + (-\ln 4 + \ln 3) + \frac{1}{12} = (\ln 2 - \ln 4) + \left(\frac{1}{6} + \frac{1}{12}\right)$$ Since $$\ln 4 = \ln 2^2 = 2 \ln 2$$, $$\ln 2 - \ln 4 = \ln 2 - 2 \ln 2 = -\ln 2$$ And $$\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$$ So, $$I = -\ln 2 + \frac{1}{4}$$ 11. **Calculate volume:** $$V = 25 \pi I = 25 \pi \left(-\ln 2 + \frac{1}{4}\right) = 25 \pi \left(\frac{1}{4} - \ln 2\right)$$ **Final answer:** $$\boxed{V = 25 \pi \left(\frac{1}{4} - \ln 2\right)}$$