Given f(x) = (2)/((x-1)(x+1)), we write it in the form:

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) = (2)/((x-1)(x+1)), we write it in the form: (2)/((x-1)(x+1)) = (A)/(x-1) + (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) = (1)/(x-1) - (1)/(x+1) The partial fraction decomposition of f(x) is (1)/(x-1) - (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. _2^7 (1)/((x+1)sqrt(x+2))\,dx = _2^3 (1)/((u^2-1)u) (2u\,du) = _2^3 (2u)/(u(u^2-1))\,du = _2^3 (2)/(u^2-1)\,du Step 4: Use the partial fraction decomposition from Part 1. From Part 1, we know that (2)/(u^2-1) = (1)/(u-1) - (1)/(u+1). Substitute this into the integral: _2^3 ((1)/(u-1) - (1)/(u+1))\,du Step 5: Integrate the expression. [|u-1| - |u+1|]_2^3 Using the logarithm property a - b = (a)/(b): = [|(u-1)/(u+1)|]_2^3 Step 6: Evaluate the definite integral using the limits. |(3-1)/(3+1)| - |(2-1)/(2+1)| = |(2)/(4)| - |(1)/(3)| = ((1)/(2)) - ((1)/(3)) Again, using the logarithm property a - b = (a)/(b): = ((1/2)/(1/3)) = ((1)/(2) × (3)/(1)) = ((3)/(2)) The value of the definite integral is (3)/(2). Got more? Send 'em!