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:

$$\frac{7-x}{(1+x)(2+x)} = \frac{8}{1+x} - \frac{9}{2+x}$$

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