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

Equating coefficients, we find \(A = 2\), \(B = -1\), and \(C = 0\).

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

(b) Integrate \(f(x)\) to find \(\int_0^4 f(x) \, dx\).

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

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

Evaluate from 0 to 4:

\(2 \ln(5) - 2 \ln(1) - \frac{1}{2} \ln(18) + \frac{1}{2} \ln(2)\).

Simplifying gives \(\ln \left( \frac{25}{3} \right)\).

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

Assume \(\frac{x^2 + 9x}{(3x - 1)(x^2 + 3)} = \frac{A}{3x - 1} + \frac{Bx + C}{x^2 + 3}\).

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

\(x^2 + 9x = A(x^2 + 3) + (Bx + C)(3x - 1)\).

Expand and collect like terms:

\(x^2 + 9x = Ax^2 + 3A + 3Bx^2 - Bx + Cx - C\).

Combine terms:

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

Equate coefficients:

\(A + 3B = 1\)

\(-B + C = 9\)

\(3A - C = 0\)

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

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

(b) Integrate \(f(x)\) from 1 to 3:

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

Integrate each term separately:

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

\(\int \frac{3}{x^2 + 3} \, dx = \sqrt{3} \arctan \left( \frac{x}{\sqrt{3}} \right)\).

Evaluate from 1 to 3:

\(\left[ \frac{1}{3} \ln |3x - 1| \right]_1^3 = \frac{1}{3} \ln 8 - \frac{1}{3} \ln 2\).

\(\left[ \sqrt{3} \arctan \left( \frac{x}{\sqrt{3}} \right) \right]_1^3 = \sqrt{3} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\sqrt{3} \pi}{6}\).

Combine results:

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

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

Assume \(\frac{15 - 6x}{(1 + 2x)(4 - x)} = \frac{A}{1 + 2x} + \frac{B}{4 - x}\).

Multiply through by \((1 + 2x)(4 - x)\) to clear the denominators:

\(15 - 6x = A(4 - x) + B(1 + 2x)\).

Expand and collect terms:

\(15 - 6x = (4A + B) + (-A + 2B)x\).

Equate coefficients:

\(4A + B = 15\)

\(-A + 2B = -6\)

Solving these equations gives \(A = 4\) and \(B = -1\).

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

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

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

Integrate each term:

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

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

Evaluate from 1 to 2:

\(2 \ln |1 + 2(2)| - 2 \ln |1 + 2(1)| - (\ln |4 - 2| - \ln |4 - 1|)\)

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

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

\(= \ln \left( \frac{5^2}{3^2 \cdot 2} \right)\)

\(= \ln \left( \frac{25}{18} \right)\)

However, the mark scheme gives the final answer as \(\ln \left( \frac{50}{27} \right)\).

(a) To express \(f(x)\) in partial fractions, assume:

\(\frac{5a}{(2x-a)(3a-x)} = \frac{A}{2x-a} + \frac{B}{3a-x}\)

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

\(5a = A(3a-x) + B(2x-a)\)

Expand and collect terms:

\(5a = (3aA - Ax) + (2xB - aB)\)

\(5a = (3aA - aB) + (2B - A)x\)

Equating coefficients, we get:

\(2B - A = 0\) and \(3aA - aB = 5a\)

From \(2B - A = 0\), we have \(A = 2B\).

Substitute \(A = 2B\) into \(3aA - aB = 5a\):

\(3a(2B) - aB = 5a\)

\(6aB - aB = 5a\)

\(5aB = 5a\)

\(B = 1\)

Thus, \(A = 2\).

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

(b) Integrate \(f(x)\):

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

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

\(= 2 \ln |2x-a| - \ln |3a-x| + C\)

Evaluate from \(a\) to \(2a\):

\(\left[ 2 \ln |2x-a| - \ln |3a-x| \right]_a^{2a}\)

\(= \left( 2 \ln |4a-a| - \ln |3a-2a| \right) - \left( 2 \ln |2a-a| - \ln |3a-a| \right)\)

\(= \left( 2 \ln 3a - \ln a \right) - \left( 2 \ln a - \ln 2a \right)\)

\(= (2 \ln 3a - \ln a) - (2 \ln a - \ln 2a)\)

\(= 2 \ln 3 - \ln 1\)

\(= \ln 9 - \ln 1\)

\(= \ln 9\)

\(= \ln 6\)

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

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

Multiply through by \((3x+2)(x^2+4)\) to clear the denominators:

\(7x + 18 = A(x^2 + 4) + (Bx + C)(3x + 2)\).

Expand and collect like terms:

\(7x + 18 = Ax^2 + 4A + 3Bx^2 + 2Bx + 3Cx + 2C\).

Combine terms: \((A + 3B)x^2 + (2B + 3C)x + (4A + 2C)\).

Equate coefficients with \(7x + 18\):

\(A + 3B = 0\)

\(2B + 3C = 7\)

\(4A + 2C = 18\)

Solve the system of equations:

From \(A + 3B = 0\), \(A = -3B\).

Substitute into \(4A + 2C = 18\):

\(4(-3B) + 2C = 18\)

\(-12B + 2C = 18\)

\(2C = 12B + 18\)

\(C = 6B + 9\)

Substitute \(C = 6B + 9\) into \(2B + 3C = 7\):

\(2B + 3(6B + 9) = 7\)

\(2B + 18B + 27 = 7\)

\(20B = -20\)

\(B = -1\)

\(A = -3(-1) = 3\)

\(C = 6(-1) + 9 = 3\)

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

(b) Find \(\int_0^2 f(x) \, dx\):

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

\(= \int_0^2 \frac{3}{3x+2} \, dx + \int_0^2 \frac{-x+3}{x^2+4} \, dx\)

First integral: \(\int \frac{3}{3x+2} \, dx = \ln|3x+2|\) evaluated from 0 to 2:

\(\left[ \ln|3(2)+2| - \ln|3(0)+2| \right] = \ln 8 - \ln 2 = \ln 4\)

Second integral: \(\int \frac{-x+3}{x^2+4} \, dx = -\frac{1}{2} \ln(x^2+4) + \frac{3}{2} \arctan\left(\frac{x}{2}\right)\) evaluated from 0 to 2:

\(\left[ -\frac{1}{2} \ln(4+4) + \frac{3}{2} \arctan(1) \right] - \left[ -\frac{1}{2} \ln(4) + \frac{3}{2} \arctan(0) \right]\)

\(= -\frac{1}{2} \ln 8 + \frac{3}{2} \frac{\pi}{4} + \frac{1}{2} \ln 4\)

\(= -\frac{1}{2} \ln 8 + \frac{3}{8} \pi + \frac{1}{2} \ln 4\)

Combine results:

\(\ln 4 - \frac{1}{2} \ln 8 + \frac{3}{8} \pi + \frac{1}{2} \ln 4\)

\(= \frac{3}{2} \ln 2 + \frac{3}{8} \pi\)