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

Assume \(f(x) = \frac{A}{2x+1} + \frac{B}{2x+3} + \frac{C}{(2x+3)^2}\).

Multiply through by \((2x+1)(2x+3)^2\) to clear the denominators:

\(10x + 9 = A(2x+3)^2 + B(2x+1)(2x+3) + C(2x+1)\).

Expand and equate coefficients to solve for \(A, B, C\):

\(A = 1, B = -1, C = 3\).

(ii) Integrate \(f(x)\) from 0 to 1:

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

Integrate each term separately:

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

\(\int \frac{1}{2x+3} \, dx = \frac{1}{2} \ln |2x+3|\),

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

Evaluate from 0 to 1:

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

Substitute limits:

\(\frac{1}{2} \ln 3 - \frac{1}{2} \ln 5 - \frac{3}{10} + \left( -\frac{1}{2} \ln 1 + \frac{1}{2} \ln 3 + \frac{3}{6} \right)\).

Simplify to get:

\(\frac{1}{2} \ln \frac{9}{5} + \frac{1}{5}\).

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

Assume \(\frac{6x^2 + 8x + 9}{(2-x)(3+2x)^2} = \frac{A}{2-x} + \frac{B}{3+2x} + \frac{C}{(3+2x)^2}\).

Multiply through by the denominator \((2-x)(3+2x)^2\) to clear the fractions:

\(6x^2 + 8x + 9 = A(3+2x)^2 + B(2-x)(3+2x) + C(2-x)\).

Expand and equate coefficients to solve for \(A, B,\) and \(C\):

\(A = 1, \; B = -1, \; C = 3\).

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

(ii) Evaluate \(\int_{-1}^{0} f(x) \, dx\):

\(\int_{-1}^{0} \left( \frac{1}{2-x} - \frac{1}{3+2x} + \frac{3}{(3+2x)^2} \right) \, dx\).

Integrate each term separately:

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

Evaluate from \(-1\) to \(0\):

\(\left[ -\ln|2-x| - \frac{1}{2} \ln|3+2x| - \frac{3}{2(3+2x)} \right]_{-1}^{0} = 1 + \frac{1}{2} \ln \left( \frac{3}{4} \right)\).

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

Assume \(\frac{5x^2 + x + 27}{(2x + 1)(x^2 + 9)} = \frac{A}{2x+1} + \frac{Bx + C}{x^2 + 9}\).

Multiply through by the denominator: \(5x^2 + x + 27 = A(x^2 + 9) + (Bx + C)(2x + 1)\).

Expand and equate coefficients:

\(5x^2 + x + 27 = Ax^2 + 9A + 2Bx^2 + Bx + Cx + C\).

\(5x^2 + x + 27 = (A + 2B)x^2 + (B + C)x + (9A + C)\).

Equating coefficients gives:

\(A + 2B = 5\)

\(B + C = 1\)

\(9A + C = 27\)

Solving these equations, we find \(A = 3\), \(B = 1\), \(C = 0\).

Thus, \(f(x) = \frac{3}{2x+1} + \frac{x}{x^2 + 9}\).

(ii) Integrate \(f(x)\):

\(\int_0^4 f(x) \, dx = \int_0^4 \left( \frac{3}{2x+1} + \frac{x}{x^2 + 9} \right) \, dx\).

Integrate each term separately:

\(\int \frac{3}{2x+1} \, dx = \frac{3}{2} \ln |2x+1|\).

\(\int \frac{x}{x^2 + 9} \, dx = \frac{1}{2} \ln |x^2 + 9|\).

Evaluate from 0 to 4:

\(\left[ \frac{3}{2} \ln |2x+1| \right]_0^4 + \left[ \frac{1}{2} \ln |x^2 + 9| \right]_0^4\).

\(= \frac{3}{2} (\ln 9 - \ln 1) + \frac{1}{2} (\ln 25 - \ln 9)\).

\(= \frac{3}{2} \ln 9 + \frac{1}{2} \ln \frac{25}{9}\).

\(= \ln 9^{3/2} + \ln \left( \frac{25}{9} \right)^{1/2}\).

\(= \ln 27 + \ln \frac{5}{3}\).

\(= \ln (27 \times \frac{5}{3})\).

\(= \ln 45\).

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

Assume \(\frac{3x^2 - 4}{x^2(3x + 2)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{3x + 2}\).

Multiply through by \(x^2(3x + 2)\) to clear the denominators:

\(3x^2 - 4 = A(x)(3x + 2) + B(3x + 2) + C(x^2)\).

Expand and collect terms:

\(3x^2 - 4 = (3A + C)x^2 + (2A + 3B)x + 2B\).

Equate coefficients:

\(3A + C = 3\)

\(2A + 3B = 0\)

\(2B = -4\)

Solve for \(B\):

\(B = -2\).

Substitute \(B = -2\) into \(2A + 3B = 0\):

\(2A - 6 = 0\)

\(A = 3\).

Substitute \(A = 3\) into \(3A + C = 3\):

\(9 + C = 3\)

\(C = -6\).

Thus, \(f(x) = \frac{3}{x} - \frac{2}{x^2} - \frac{6}{3x+2}\).

(ii) Integrate \(f(x)\) from 1 to 2:

\(\int_1^2 \left( \frac{3}{x} - \frac{2}{x^2} - \frac{6}{3x+2} \right) \, dx\).

Integrate each term separately:

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

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

\(\int \frac{-6}{3x+2} \, dx = -2 \ln |3x+2|\)

Evaluate from 1 to 2:

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

\(= \left( 3 \ln 2 + 1 - 2 \ln 8 \right) - \left( 0 + 2 - 2 \ln 5 \right)\)

\(= 3 \ln 2 - 2 \ln 8 - 2 + 2 \ln 5\)

\(= \ln\left(\frac{8}{8}\right) - 1\)

\(= \ln\left(\frac{25}{8}\right) - 1\).

(i) To show that \((x + 1)\) is a factor of \(4x^3 - x^2 - 11x - 6\), substitute \(x = -1\) into the polynomial:

\(4(-1)^3 - (-1)^2 - 11(-1) - 6 = -4 - 1 + 11 - 6 = 0\).

Since the result is 0, \((x + 1)\) is a factor.

(ii) To find \(\int \frac{4x^2 + 9x - 1}{4x^3 - x^2 - 11x - 6} \, dx\), perform polynomial division of \(4x^3 - x^2 - 11x - 6\) by \(x + 1\) to obtain the quotient \(4x^2 - 5x - 6\).

Express the integrand as partial fractions:

\(\frac{4x^2 + 9x - 1}{4x^3 - x^2 - 11x - 6} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{4x+3}\).

Solving for constants, we find \(A = -2\), \(B = 1\), \(C = 8\).

Integrate each term:

\(-2 \int \frac{1}{x+1} \, dx + \int \frac{1}{x-2} \, dx + 2 \int \frac{1}{4x+3} \, dx\).

The integrals yield:

\(-2 \ln|x+1| + \ln|x-2| + 2 \ln|4x+3|\).

Thus, the solution is \(-2 \ln(x+1) + \ln(x-2) + 2 \ln(4x+3)\).