(i) Express \(f(x)\) as:

\(f(x) = \frac{A}{2x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}\)

Multiply through by \((2x-1)(x+2)^2\) to clear the denominators:

\(11x + 7 = A(x+2)^2 + B(2x-1)(x+2) + C(2x-1)\)

Expand and equate coefficients to find \(A = 2\), \(B = -1\), \(C = 3\).

Thus, \(f(x) = \frac{2}{2x-1} - \frac{1}{x+2} + \frac{3}{(x+2)^2}\).

(ii) Integrate each term separately:

\(\int \frac{2}{2x-1} \, dx = \ln|2x-1|\)

\(\int -\frac{1}{x+2} \, dx = -\ln|x+2|\)

\(\int \frac{3}{(x+2)^2} \, dx = -\frac{3}{x+2}\)

Evaluate from 1 to 2:

\(\left[ \ln|2x-1| - \ln|x+2| - \frac{3}{x+2} \right]_1^2\)

\(= \left( \ln(3) - \ln(4) - \frac{3}{4} \right) - \left( \ln(1) - \ln(3) - \frac{3}{3} \right)\)

\(= \ln\left(\frac{3}{4}\right) + 1 - \left(0 - \ln(3) - 1\right)\)

\(= \ln\left(\frac{3}{4}\right) + 1 + \ln(3) + 1\)

\(= \ln\left(\frac{9}{4}\right) + \frac{1}{4}\)

(i) To express \(f(x)\) in the form \(\frac{A}{2-x} + \frac{Bx+C}{2+x^2}\), we equate:

\(\frac{6 + 6x}{(2-x)(2+x^2)} = \frac{A}{2-x} + \frac{Bx+C}{2+x^2}\)

Multiply through by \((2-x)(2+x^2)\) to clear the denominators:

\(6 + 6x = A(2+x^2) + (Bx+C)(2-x)\)

Expand and collect like terms:

\(6 + 6x = 2A + Ax^2 + 2Bx + Cx - Bx^2 - Cx\)

\(6 + 6x = (A-B)x^2 + (2B+C)x + 2A\)

Equating coefficients, we get:

\(A - B = 0\)

\(2B + C = 6\)

\(2A = 6\)

Solving these equations, we find:

\(A = 3\)

\(B = 3\)

\(C = 0\)

(ii) To show \(\int_{-1}^{1} f(x) \, dx = 3 \ln 3\), integrate each term separately:

\(\int \frac{3}{2-x} \, dx = -3 \ln |2-x|\)

\(\int \frac{3x}{2+x^2} \, dx = \frac{3}{2} \ln |2+x^2|\)

\(\int \frac{0}{2+x^2} \, dx = 0\)

Evaluate from \(-1\) to \(1\):

\(-3 \ln |2-1| + \frac{3}{2} \ln |2+1^2| - (-3 \ln |2+1| + \frac{3}{2} \ln |2+(-1)^2|)\)

\(= -3 \ln 1 + \frac{3}{2} \ln 3 + 3 \ln 3 - \frac{3}{2} \ln 3\)

\(= 3 \ln 3\)

(a) Express \(f(x)\) in partial fractions:

Assume \(f(x) = \frac{A}{2x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^2}\).

Multiply through by \((2x+1)(x+2)^2\) to clear the denominators:

\(3 - 3x^2 = A(x+2)^2 + B(2x+1)(x+2) + C(2x+1)\).

Expand and equate coefficients to find \(A = 1\), \(B = -2\), \(C = 3\).

Thus, \(f(x) = \frac{1}{2x+1} - \frac{2}{x+2} + \frac{3}{(x+2)^2}\).

(b) Integrate \(f(x)\) from 0 to 4:

\(\int_0^4 f(x) \, dx = \int_0^4 \left( \frac{1}{2x+1} - \frac{2}{x+2} + \frac{3}{(x+2)^2} \right) \, dx\).

Integrate each term separately:

\(\int \frac{1}{2x+1} \, dx = \frac{1}{2} \ln |2x+1|\),

\(\int \frac{-2}{x+2} \, dx = -2 \ln |x+2|\),

\(\int \frac{3}{(x+2)^2} \, dx = \frac{-3}{x+2}\).

Evaluate from 0 to 4:

\(\left[ \frac{1}{2} \ln |2x+1| - 2 \ln |x+2| - \frac{3}{x+2} \right]_0^4\).

Substitute limits:

\(\left( \frac{1}{2} \ln 9 - 2 \ln 6 - \frac{3}{6} \right) - \left( \frac{1}{2} \ln 1 - 2 \ln 2 - \frac{3}{2} \right)\).

Simplify to obtain \(1 - \ln 3\).

To express \(\frac{7x^2 - 3x + 2}{x(x^2 + 1)}\) in partial fractions, we assume the form:

\(\frac{A}{x} + \frac{Bx + C}{x^2 + 1}\)

Multiply through by the denominator \(x(x^2 + 1)\) to clear the fractions:

\(7x^2 - 3x + 2 = A(x^2 + 1) + (Bx + C)x\)

Expanding the right side gives:

\(7x^2 - 3x + 2 = Ax^2 + A + Bx^2 + Cx\)

Combine like terms:

\(7x^2 - 3x + 2 = (A + B)x^2 + Cx + A\)

Equate coefficients:

1. \(A + B = 7\)

2. \(C = -3\)

3. \(A = 2\)

From equation 3, \(A = 2\).

Substitute \(A = 2\) into equation 1:

\(2 + B = 7\)

\(B = 5\)

Thus, the partial fraction decomposition is:

\(\frac{2}{x} + \frac{5x - 3}{x^2 + 1}\)

(i) Let \(u = \sqrt{6-x}\), then \(u^2 = 6-x\) and differentiating gives \(2u \, du = -dx\), or \(dx = -2u \, du\).

Substitute \(x = 6-u^2\) into the integral:

\(I = \int \frac{5}{6-u^2 + u} (-2u) \, du\).

Simplify the integrand:

\(I = \int \frac{-10u}{6-u^2+u} \, du = \int \frac{-10u}{(3-u)(2+u)} \, du\).

Change the limits: when \(x = 2\), \(u = \sqrt{4} = 2\); when \(x = 5\), \(u = \sqrt{1} = 1\).

Thus, \(I = \int_{1}^{2} \frac{10u}{(3-u)(2+u)} \, du\).

(ii) Use partial fraction decomposition:

\(\frac{10u}{(3-u)(2+u)} = \frac{A}{3-u} + \frac{B}{2+u}\).

Multiply through by \((3-u)(2+u)\) and equate coefficients to find \(A = 6\) and \(B = -4\).

Integrate each term separately:

\(\int \frac{6}{3-u} \, du = -6 \ln|3-u|\) and \(\int \frac{-4}{2+u} \, du = -4 \ln|2+u|\).

Combine and evaluate from 1 to 2:

\(I = [-6 \ln|3-u| - 4 \ln|2+u|]_{1}^{2}\).

Calculate:

\(I = [-6 \ln(1) - 4 \ln(4)] - [-6 \ln(2) - 4 \ln(3)]\).

Simplify to find \(I = 2 \ln\left(\frac{9}{2}\right)\).