(i) Express \(f(x)\) in the form \(A + \frac{B}{2x+1} + \frac{C}{x+2}\).

Equating coefficients, we find \(A = 2\), \(B = 1\), and \(C = -2\).

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

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

\(\int f(x) \, dx = \int \left( 2 + \frac{1}{2x+1} - \frac{2}{x+2} \right) \, dx\)

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

Evaluate from 0 to 4:

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

= \(\left( 8 + \frac{1}{2} \ln 9 - 2 \ln 6 \right) - \left( 0 + \frac{1}{2} \ln 1 - 2 \ln 2 \right)\)

= \(8 + \frac{1}{2} \ln 9 - 2 \ln 6 + 2 \ln 2\)

= \(8 + \frac{1}{2} \ln 9 - 2 \ln 3\)

= \(8 - \ln 3\).

(i) To express \(f(x)\) in the form \(A + \frac{B}{x} + \frac{Cx + D}{x^2 + 2}\), perform partial fraction decomposition:

1. Set \(f(x) = \frac{3x^3 + 6x - 8}{x(x^2 + 2)} = A + \frac{B}{x} + \frac{Cx + D}{x^2 + 2}\).

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

\(3x^3 + 6x - 8 = A(x^2 + 2) + B(x^2 + 2) + (Cx + D)x\).

3. Expand and collect like terms:

\(3x^3 + 6x - 8 = Ax^2 + 2A + Bx^2 + 2B + Cx^2 + Dx\).

4. Equate coefficients:

\(A + B + C = 3\)

\(D = 6\)

\(2A + 2B = -8\)

5. Solve the system of equations to find \(A = 3, B = -4, C = 4, D = 0\).

(ii) To show \(\int_1^2 f(x) \, dx = 3 - \ln 4\):

1. Integrate \(f(x) = 3 - \frac{4}{x} + \frac{4x}{x^2 + 2}\).

2. Integrate each term separately:

\(\int 3 \, dx = 3x\)

\(\int -\frac{4}{x} \, dx = -4 \ln |x|\)

\(\int \frac{4x}{x^2 + 2} \, dx = 2 \ln(x^2 + 2)\)

3. Evaluate from 1 to 2:

\(\left[ 3x - 4 \ln |x| + 2 \ln(x^2 + 2) \right]_1^2\)

4. Substitute the limits:

\((6 - 4 \ln 2 + 2 \ln 6) - (3 - 4 \ln 1 + 2 \ln 3)\)

5. Simplify to obtain \(3 - \ln 4\).

Let \(u = e^x\), then \(du = e^x \, dx\) or \(dx = \frac{du}{u}\).

The limits of integration change from \(x = 0\) to \(x = \ln 4\), which correspond to \(u = 1\) to \(u = 4\).

Substitute into the integral:

\(\int_1^4 \frac{u^2}{u^2 + 3u + 2} \, \frac{du}{u} = \int_1^4 \frac{u}{u^2 + 3u + 2} \, du.\)

Factor the denominator: \(u^2 + 3u + 2 = (u+1)(u+2)\).

Use partial fraction decomposition:

\(\frac{u}{(u+1)(u+2)} = \frac{A}{u+1} + \frac{B}{u+2}.\)

Multiply through by the denominator:

\(u = A(u+2) + B(u+1).\)

Expanding gives:

\(u = Au + 2A + Bu + B.\)

Combine like terms:

\(u = (A+B)u + (2A+B).\)

Equate coefficients:

\(A + B = 1\) and \(2A + B = 0\).

Solving these equations gives \(A = 2\) and \(B = -1\).

Thus, \(\frac{u}{(u+1)(u+2)} = \frac{2}{u+2} - \frac{1}{u+1}\).

Integrate each term:

\(\int_1^4 \left( \frac{2}{u+2} - \frac{1}{u+1} \right) \, du = 2 \ln|u+2| - \ln|u+1| \bigg|_1^4.\)

Evaluate the definite integral:

\(\left( 2 \ln 6 - \ln 5 \right) - \left( 2 \ln 3 - \ln 2 \right).\)

Simplify using logarithm properties:

\(2 \ln 6 - \ln 5 - 2 \ln 3 + \ln 2 = \ln \left( \frac{36}{5} \right) - \ln \left( \frac{9}{2} \right).\)

Combine the logarithms:

\(\ln \left( \frac{36}{5} \times \frac{2}{9} \right) = \ln \left( \frac{8}{5} \right).\)

(i) Express \(f(x)\) in the form \(A + \frac{B}{x+1} + \frac{C}{2x-3}\).

Equating coefficients, we find:

\(A = 2\), \(B = -2\), \(C = -1\).

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

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

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

This becomes:

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

Evaluate from 2 to 6:

\(\left[2x - 2\ln|x+1| - \frac{1}{2}\ln|2x-3|\right]_2^6\).

Calculate:

At \(x = 6\): \(12 - 2\ln 7 - \frac{1}{2}\ln 9\).

At \(x = 2\): \(4 - 2\ln 3 - \frac{1}{2}\ln 1\).

Result: \(8 - \ln\left(\frac{49}{3}\right)\).

First, express \(\frac{4x^2 + 5x + 3}{2x^2 + 5x + 2}\) in partial fractions:

Assume \(\frac{4x^2 + 5x + 3}{2x^2 + 5x + 2} = A + \frac{B}{2x+1} + \frac{C}{x+2}\).

Equating coefficients, solve for \(A, B,\) and \(C\):

\(A = 2, \; B = 1, \; C = -3\).

Thus, \(\frac{4x^2 + 5x + 3}{2x^2 + 5x + 2} = 2 + \frac{1}{2x+1} - \frac{3}{x+2}\).

Now integrate each term separately:

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

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

Calculate the definite integral:

\(= \left( 8 + \frac{1}{2} \ln 9 - 3 \ln 6 \right) - \left( 0 + \frac{1}{2} \ln 1 - 3 \ln 2 \right)\).

\(= 8 + \frac{1}{2} \ln 9 - 3 \ln 6 + 3 \ln 2\).

Simplify the expression:

\(= 8 + \frac{1}{2} \ln 9 - 3 \ln 3 - 3 \ln 2 + 3 \ln 2\).

\(= 8 - \ln 9\).