Answer
$$A = \frac{3}{2} + 4\ln \frac{3}{4}$$
Work Step by Step
$$\eqalign{
& 0 \geqslant \frac{{x - 3}}{{{x^2} + {x^3}}}{\text{ on the interval }}\left[ {1,2} \right],{\text{ then}} \cr
& {\text{The area is given by }} \cr
& A = \int_1^2 {\frac{{3 - x}}{{{x^2} + {x^3}}}} dx \cr
& A = \int_1^2 {\frac{{x - 3}}{{{x^2}\left( {x + 1} \right)}}} dx \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{{3 - x}}{{{x^2}\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x + 1}} \cr
& 3 - x = Ax\left( {x + 1} \right) + B\left( {x + 1} \right) + C{x^2} \cr
& 3 - x = A{x^2} + Ax + Bx + B + C{x^2} \cr
& 3 - x = \left( {A{x^2} + C{x^2}} \right) + \left( {Ax + Bx} \right) + B \cr
& 3 - x = \left( {A + C} \right){x^2} + \left( {A + B} \right)x + B \cr
& B = 3,{\text{ }}A + B = 1 \Rightarrow A = - 4,{\text{ }}C = 4 \cr
& \frac{{3 - x}}{{{x^2}\left( {x + 1} \right)}} = - \frac{4}{x} + \frac{3}{{{x^2}}} + \frac{4}{{x + 1}} \cr
& {\text{Therefore}} \cr
& A = \int_1^2 {\left( { - \frac{4}{x} + \frac{3}{{{x^2}}} + \frac{4}{{x + 1}}} \right)} dx \cr
& A = \left[ { - 4\ln \left| x \right| - \frac{3}{x} - 4\ln \left| {x + 1} \right|} \right]_1^2 \cr
& A = \left[ { - 4\ln \left| 2 \right| - \frac{3}{2} - 4\ln \left| {2 + 1} \right|} \right] - \left[ { - 4\ln \left| 1 \right| - \frac{3}{1} - 4\ln \left| {1 + 1} \right|} \right] \cr
& A = \left[ { - 4\ln \left| 2 \right| - \frac{3}{2} - 4\ln \left| 3 \right|} \right] - \left[ { - 4\ln \left| 1 \right| - \frac{3}{1} - 4\ln \left| 2 \right|} \right] \cr
& A = - 4\ln \left| 2 \right| - \frac{3}{2} - 4\ln \left| 3 \right| + 3 + 4\ln \left| 2 \right| \cr
& A = \frac{3}{2} + 4\ln \frac{3}{4} \cr} $$
