To express \(\frac{4x^2 - 13x + 13}{(2x - 1)(x - 3)}\) in partial fractions, we start by assuming:

\(\frac{4x^2 - 13x + 13}{(2x - 1)(x - 3)} = \frac{A}{2x - 1} + \frac{B}{x - 3}\)

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

\(4x^2 - 13x + 13 = A(x - 3) + B(2x - 1)\)

Expanding both sides gives:

\(4x^2 - 13x + 13 = Ax - 3A + 2Bx - B\)

Combine like terms:

\(4x^2 - 13x + 13 = (A + 2B)x - (3A + B)\)

Equating coefficients, we get:

\(A + 2B = -13\)

\(-3A - B = 13\)

Solving these equations simultaneously, we find:

\(A = 2\), \(B = -3\)

Thus, the partial fraction decomposition is:

\(\frac{4x^2 - 13x + 13}{(2x - 1)(x - 3)} = \frac{2}{2x - 1} + \frac{-3}{x - 3}\)

Alternatively, dividing the numerator by the denominator gives:

\(2 + \frac{-x + 7}{(2x - 1)(x - 3)}\)

(i) Express \(f(x)\) as a sum of partial fractions:

\(\frac{x^2 + 3x + 3}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}\)

Multiply through by \((x+1)(x+3)\):

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

Expand and collect terms:

\(x^2 + 3x + 3 = (A+B)x + (3A + B)\)

Equate coefficients:

\(A + B = 1\)

\(3A + B = 3\)

Solve these equations to find \(A = 1\), \(B = \frac{1}{2}\), \(C = -\frac{3}{2}\).

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

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

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

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

Evaluate from 0 to 3:

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

\(= \left( 3 + \frac{1}{2} \ln 4 - \frac{3}{2} \ln 6 \right) - \left( 0 + \frac{1}{2} \ln 1 - \frac{3}{2} \ln 3 \right)\)

\(= 3 + \frac{1}{2} \ln 4 - \frac{3}{2} \ln 6 + \frac{3}{2} \ln 3\)

\(= 3 + \frac{1}{2} \ln \left( \frac{4 \times 3}{6} \right)\)

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

Thus, \(\int_0^3 f(x) \, dx = 3 - \frac{1}{2} \ln 2\).

(i) To express \(f(x)\) in the given form, perform polynomial long division on \(\frac{x^3 - x - 2}{(x-1)(x^2+1)}\) to find \(A\), and then equate the numerator to find \(B, C,\) and \(D\).

Divide \(x^3 - x - 2\) by \((x-1)(x^2+1)\) to get a quotient of \(A = 1\) and a remainder.

Equate the remainder to \(B(x^2+1) + (Cx+D)(x-1)\) and solve for \(B, C,\) and \(D\):

\(B = -1, C = 2, D = 0\).

(ii) Integrate \(f(x) = 1 - \frac{1}{x-1} + \frac{2x}{x^2+1}\) from 2 to 3:

\(\int_2^3 \left( 1 - \frac{1}{x-1} + \frac{2x}{x^2+1} \right) \, dx = \left[ x - \ln|x-1| + \ln(x^2+1) \right]_2^3\).

Evaluate the definite integral:

\(\left[ 3 - \ln|3-1| + \ln(3^2+1) \right] - \left[ 2 - \ln|2-1| + \ln(2^2+1) \right] = 1\).

(i) To express \(f(x)\) in the form \(A + \frac{B}{x+2} + \frac{C}{2x-1}\), perform partial fraction decomposition:

\(\frac{4x^2 + 9x - 8}{(x+2)(2x-1)} = A + \frac{B}{x+2} + \frac{C}{2x-1}\)

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

\(4x^2 + 9x - 8 = A(x+2)(2x-1) + B(2x-1) + C(x+2)\)

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

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

(ii) Integrate \(f(x)\) using the partial fraction decomposition:

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

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

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

Evaluate from 1 to 4:

\(\left[ 2x + 2 \ln |x+2| - \frac{1}{2} \ln |2x-1| \right]_1^4\)

\(= \left( 8 + 2 \ln 6 - \frac{1}{2} \ln 7 \right) - \left( 2 + 2 \ln 3 - \frac{1}{2} \ln 1 \right)\)

\(= 6 + 2 \ln \frac{6}{3} - \frac{1}{2} \ln \frac{7}{1}\)

\(= 6 + \ln 2 - \frac{1}{2} \ln 7\)

\(= 6 + \frac{1}{2} \ln \left( \frac{16}{7} \right)\)

First, express the integrand \(\frac{u-1}{u+1}\) in the form \(A + \frac{B}{u+1}\).

We have:

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

Thus, \(A = 1\) and \(B = -2\).

Now, integrate:

\(\int_1^2 \left( 1 - \frac{2}{u+1} \right) \, du = \int_1^2 1 \, du - 2 \int_1^2 \frac{1}{u+1} \, du\)

The first integral is:

\(\int_1^2 1 \, du = [u]_1^2 = 2 - 1 = 1\)

The second integral is:

\(-2 \int_1^2 \frac{1}{u+1} \, du = -2 [\ln|u+1|]_1^2\)

Evaluate the limits:

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

Combine the results:

\(1 - 2 \ln \frac{3}{2} = 1 + \ln \left( \frac{1}{\left(\frac{3}{2}\right)^2} \right) = 1 + \ln \frac{4}{9}\)