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

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

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

\(7x + 4 = A(x+1)^2 + B(2x+1)(x+1) + C(2x+1)\).

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

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

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

(ii) Integrate \(f(x)\) from 0 to 2:

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

Integrate each term separately:

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

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

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

Evaluate from 0 to 2:

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

Calculate the definite integral:

At \(x = 2\): \(\ln 5 - \ln 3 - \frac{3}{3} = \ln \frac{5}{3} - 1\).

At \(x = 0\): \(\ln 1 - \ln 1 - 3 = -3\).

Result: \((\ln \frac{5}{3} - 1) - (-3) = \ln \frac{5}{3} + 2\).

Thus, \(\int_0^2 f(x) \, dx = 2 + \ln \frac{5}{3}\).

(a)(i) The expression \(\frac{4x}{(x+4)(x^2+3)}\) can be decomposed into partial fractions as \(\frac{A}{x+4} + \frac{Bx+C}{x^2+3}\) because \(x^2+3\) is an irreducible quadratic factor.

(a)(ii) The expression \(\frac{2x+1}{(x-2)(x+2)^2}\) can be decomposed into partial fractions as \(\frac{A}{x-2} + \frac{Bx+C}{(x+2)^2}\) because \((x+2)^2\) is a repeated linear factor.

(b) To show \(\int_3^4 \frac{3x}{(x+1)(x-2)} \, dx = \ln 5\), first express \(\frac{3x}{(x+1)(x-2)}\) as \(\frac{A}{x+1} + \frac{B}{x-2}\). Solving for \(A\) and \(B\), we find \(A = 1\) and \(B = 2\).

Integrate: \(\int \left( \frac{1}{x+1} + \frac{2}{x-2} \right) \, dx = \ln|x+1| + 2\ln|x-2| + C\).

Evaluate from 3 to 4: \(\left[ \ln|x+1| + 2\ln|x-2| \right]_3^4 = (\ln 5 + 2\ln 2) - (\ln 4 + 2\ln 1) = \ln 5\).

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

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

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

\(4x = A(x+1)^2 + B(3x+1) + C(3x+1)(x+1)\).

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

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

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

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

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

Integrate each term separately:

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

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

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

Evaluate from 0 to 1:

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

\(= -\ln 4 - 1 + \ln 2 + 2\).

\(= 1 - \ln 2\).

(a) Express \(f(x)\) in the form \(\frac{Ax + B}{4 + x^2} + \frac{C}{1 + x}\).

Equating coefficients, we have:

\((Ax + B)(1 + x) + C(4 + x^2) = 5x^2 + x + 11\).

Solving, we find \(A = 2\), \(B = -1\), \(C = 3\).

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

(b) Integrate each term separately:

\(\int \frac{2x - 1}{4 + x^2} \, dx = \frac{A}{2} \ln(4 + x^2) + \frac{B}{2} \arctan\left(\frac{x}{2}\right)\).

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

Substitute limits 0 and 2:

\(a \ln(4 + 4) + b \arctan\left(\frac{2}{2}\right) + c \ln(1 + 2) - [a \ln(4) + b \arctan(0) + c \ln(1)]\).

Calculate: \(\ln 54 - \frac{\pi}{8}\).

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

Equate coefficients to find \(A = 2\), \(B = 0\), and \(C = 1\).

(b) Integrate to obtain \(-2 \ln(3-x)\) and \(\frac{1}{\sqrt{3}} \arctan(\sqrt{3}x)\).

Substitute limits to find \(2 \ln \frac{3}{2} + \frac{1}{\sqrt{3}} \pi\).