You're on a roll — Here is the solution for both parts of the problem. Part 1: Express $f(x)$ in partial fractions. Step 1: Set up the partial fraction decomposition. Given $f(x) = \frac{2}{(x-1)(x+1)}$, we write it in the form: $$ \frac{2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} $$ Step 2: Clear the denominators. Multiply both sides by $(x-1)(x+1)$: $$ 2 = A(x+1) + B(x-1) $$ Step 3: Solve for the constants $A$ and $B$. To find $A$, set $x=1$: $$ 2 = A(1+1) + B(1-1) $$ $$ 2 = 2A $$ $$ A = 1 $$ To find $B$, set $x=-1$: $$ 2 = A(-1+1) + B(-1-1) $$ $$ 2 = -2B $$ $$ B = -1 $$ Step 4: Write the final partial fraction expression. Substitute the values of $A$ and $B$ back into the decomposition: $$ f(x) = \frac{1}{x-1} - \frac{1}{x+1} $$ The partial fraction decomposition of $f(x)$ is $\boxed{\frac{1}{x-1} - \frac{1}{x+1}}$. Part 2: Evaluate the definite integral. Step 1: Apply the substitution. Let $u = \sqrt{x+2}$. Then $u^2 = x+2$, which implies $x = u^2 - 2$. From this, $x+1 = (u^2 - 2) + 1 = u^2 - 1$. Differentiating $u^2 = x+2$ with respect to $x$, we get $2u\,du = dx$. Step 2: Change the limits of integration. When $x=2$, $u = \sqrt{2+2} = \sqrt{4} = 2$. When $x=7$, $u = \sqrt{7+2} = \sqrt{9} = 3$. Step 3: Rewrite the integral in terms of $u$. $$ \int_{2}^{7} \frac{1}{(x+1)\sqrt{x+2}}\,dx = \int_{2}^{3} \frac{1}{(u^2-1)u} (2u\,du) $$ $$ = \int_{2}^{3} \frac{2u}{u(u^2-1)}\,du $$ $$ = \int_{2}^{3} \frac{2}{u^2-1}\,du $$ Step 4: Use the partial fraction decomposition from Part 1. From Part 1, we know that $\frac{2}{u^2-1} = \frac{1}{u-1} - \frac{1}{u+1}$. Substitute this into the integral: $$ \int_{2}^{3} \left(\frac{1}{u-1} - \frac{1}{u+1}\right)\,du $$ Step 5: Integrate the expression. $$ \left[\ln|u-1| - \ln|u+1|\right]_{2}^{3} $$ Using the logarithm property $\ln a - \ln b = \ln \frac{a}{b}$: $$ = \left[\ln\left|\frac{u-1}{u+1}\right|\right]_{2}^{3} $$ Step 6: Evaluate the definite integral using the limits. $$ \ln\left|\frac{3-1}{3+1}\right| - \ln\left|\frac{2-1}{2+1}\right| $$ $$ = \ln\left|\frac{2}{4}\right| - \ln\left|\frac{1}{3}\right| $$ $$ = \ln\left(\frac{1}{2}\right) - \ln\left(\frac{1}{3}\right) $$ Again, using the logarithm property $\ln a - \ln b = \ln \frac{a}{b}$: $$ = \ln\left(\frac{1/2}{1/3}\right) $$ $$ = \ln\left(\frac{1}{2} \times \frac{3}{1}\right) $$ $$ = \ln\left(\frac{3}{2}\right) $$ The value of the definite integral is $\boxed{\ln\frac{3}{2}}$. Got more? Send 'em!