Step 1: Express the fraction in partial fractions. Let $\frac{7-x}{(1+x)(2+x)} = \frac{A}{1+x} + \frac{B}{2+x}$. Multiply both sides by $(1+x)(2+x)$: $$ 7-x = A(2+x) + B(1+x) $$ To find $A$, set $x = -1$: $$ 7 - (-1) = A(2 - 1) + B(1 - 1) $$ $$ 8 = A(1) + B(0) $$ $$ A = 8 $$ To find $B$, set $x = -2$: $$ 7 - (-2) = A(2 - 2) + B(1 - 2) $$ $$ 9 = A(0) + B(-1) $$ $$ 9 = -B $$ $$ B = -9 $$ So, the partial fraction decomposition is: $$ \frac{7-x}{(1+x)(2+x)} = \frac{8}{1+x} - \frac{9}{2+x} $$ Step 2: Integrate the partial fractions. We need to evaluate $\int \frac{7-x}{(1+x)(2+x)} dx$. Using the partial fractions from Step 1: $$ \int \left( \frac{8}{1+x} - \frac{9}{2+x} \right) dx $$ Integrate each term: $$ \int \frac{8}{1+x} dx = 8 \ln|1+x| $$ $$ \int \frac{9}{2+x} dx = 9 \ln|2+x| $$ Combine the results: $$ \int \frac{7-x}{(1+x)(2+x)} dx = 8 \ln|1+x| - 9 \ln|2+x| + C $$ Using logarithm properties ($a \ln b = \ln b^a$ and $\ln a - \ln b = \ln \frac{a}{b}$): $$ 8 \ln|1+x| - 9 \ln|2+x| + C = \ln|(1+x)^8| - \ln|(2+x)^9| + C $$ $$ = \ln \left| \frac{(1+x)^8}{(2+x)^9} \right| + C $$ The question asks to show that the integral is $\ln \left| \frac{(1+x)^9