To find the constants \(A, B, C,\) and \(D\), we start by expressing \(\frac{2x^3 - 1}{x^2(2x-1)}\) as a sum of partial fractions:

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

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

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

Expanding and equating coefficients, we find:

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

For the integral, substitute the partial fraction decomposition:

\(\int_1^2 \left( 1 + \frac{2}{x} + \frac{1}{x^2} - \frac{3}{2x-1} \right) \, dx.\)

Integrate term by term:

\(\int 1 \, dx = x\)

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

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

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

Evaluate from 1 to 2:

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

Substitute the limits:

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

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

Calculate the difference:

\(2 + 2 \ln 2 - \frac{1}{2} - \frac{3}{2} \ln 3 - (1 - 1) = \frac{3}{2} + \frac{1}{2} \ln \left( \frac{16}{27} \right)\)

(a) Express \(f(x)\) in the form:

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

Using partial fraction decomposition, equate:

\(2x^2 + 17x - 17 = A(2-x)^2 + B(1+2x)(2-x) + C(1+2x)\)

Expand and compare coefficients to find:

\(A = -4\), \(B = -3\), \(C = 5\).

(b) Integrate each term separately:

\(\int \frac{-4}{1+2x} \, dx = -2 \ln(1+2x)\)

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

\(\int \frac{5}{(2-x)^2} \, dx = \frac{5}{2-x}\)

Substitute limits from 0 to 1:

\(-2 \ln(3) + 3 \ln(1) + \frac{5}{1} - (-2 \ln(1) + 3 \ln(2) + 5/2)\)

Simplify to obtain:

\(\frac{5}{2} - \ln 72\).

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

Assume \(\frac{2}{(2x-1)(2x+1)} = \frac{A}{2x-1} + \frac{B}{2x+1}\).

Multiply through by \((2x-1)(2x+1)\) to get:

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

Expanding gives \(2 = (2A + 2B)x + (A - B)\).

Equating coefficients, we have:

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

Solving these equations, we find \(A = 1\) and \(B = -1\).

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

(b) Square the result of part (a):

\((f(x))^2 = \left( \frac{1}{2x-1} - \frac{1}{2x+1} \right)^2\).

Expanding gives:

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

Using \(f(x) = \frac{1}{2x-1} - \frac{1}{2x+1}\), we have:

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

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

(c) Integrate \((f(x))^2\) from 1 to 2:

\(\int_1^2 (f(x))^2 \, dx = \int_1^2 \left( \frac{1}{(2x-1)^2} - \frac{1}{2x-1} + \frac{1}{2x+1} + \frac{1}{(2x+1)^2} \right) \, dx\).

Integrate each term separately:

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

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

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

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

Evaluate from 1 to 2:

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

Substitute limits and simplify to obtain:

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

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

Assume \(\frac{2x^2 + x + 8}{(2x - 1)(x^2 + 2)} = \frac{A}{2x-1} + \frac{Bx + C}{x^2 + 2}\).

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

\(2x^2 + x + 8 = A(x^2 + 2) + (Bx + C)(2x - 1)\).

Expand and equate coefficients:

\(2x^2 + x + 8 = Ax^2 + 2A + 2Bx^2 - Bx + 2Cx - C\).

Combine like terms:

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

Equate coefficients:

\(A + 2B = 2\)

\(-B + 2C = 1\)

\(2A - C = 8\)

Solve the system of equations:

From \(A + 2B = 2\), \(A = 2 - 2B\).

Substitute into \(2A - C = 8\):

\(2(2 - 2B) - C = 8\)

\(4 - 4B - C = 8\)

\(C = -4B - 4\)

Substitute \(C = -4B - 4\) into \(-B + 2C = 1\):

\(-B + 2(-4B - 4) = 1\)

\(-B - 8B - 8 = 1\)

\(-9B = 9\)

\(B = -1\)

Substitute \(B = -1\) into \(A = 2 - 2B\):

\(A = 2 - 2(-1) = 4\)

Substitute \(B = -1\) into \(C = -4B - 4\):

\(C = -4(-1) - 4 = 0\)

Thus, \(A = 4\), \(B = -1\), \(C = 0\).

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

(ii) Integrate \(f(x)\) from 1 to 5:

\(\int_1^5 \left( \frac{4}{2x-1} - \frac{x}{x^2+2} \right) \, dx\).

Integrate each term separately:

\(\int \frac{4}{2x-1} \, dx = 2 \ln |2x-1|\)

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

Evaluate from 1 to 5:

\(\left[ 2 \ln |2x-1| - \frac{1}{2} \ln |x^2+2| \right]_1^5\)

\(= \left( 2 \ln 9 - \frac{1}{2} \ln 27 \right) - \left( 2 \ln 1 - \frac{1}{2} \ln 3 \right)\)

\(= 2 \ln 9 - \frac{1}{2} \ln 27 + \frac{1}{2} \ln 3\)

\(= \ln 81 - \ln 27 + \ln 3\)

\(= \ln \left( \frac{81 \times 3}{27} \right)\)

\(= \ln 27\)

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

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

Multiply through by the denominator \(x^2(x+2)\) to get:

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

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

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

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

(ii) Integrate \(f(x)\) from 1 to 4:

\(\int_1^4 \left( \frac{-1}{x} + \frac{3}{x^2} + \frac{2}{x+2} \right) \, dx\).

Integrate each term separately:

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

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

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

Evaluate from 1 to 4:

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

Substitute the limits:

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

Simplify:

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

Combine logs:

\(-\ln 4 + 2\ln 2 + 3 - \frac{3}{4}\).

\(= \frac{9}{4}\).