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:

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

(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:

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

\(= 8 - \ln 9\).

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:

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

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

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

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

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

\(10x + 9 = A(2x+3)^2 + B(2x+1)(2x+3) + C(2x+1)\).

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

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

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

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

Integrate each term separately:

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

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

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

Evaluate from 0 to 1:

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

Substitute limits:

\(\frac{1}{2} \ln 3 - \frac{1}{2} \ln 5 - \frac{3}{10} + \left( -\frac{1}{2} \ln 1 + \frac{1}{2} \ln 3 + \frac{3}{6} \right)\).

Simplify to get:

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

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

Assume \(\frac{6x^2 + 8x + 9}{(2-x)(3+2x)^2} = \frac{A}{2-x} + \frac{B}{3+2x} + \frac{C}{(3+2x)^2}\).

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

\(6x^2 + 8x + 9 = A(3+2x)^2 + B(2-x)(3+2x) + C(2-x)\).

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

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

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

(ii) Evaluate \(\int_{-1}^{0} f(x) \, dx\):

\(\int_{-1}^{0} \left( \frac{1}{2-x} - \frac{1}{3+2x} + \frac{3}{(3+2x)^2} \right) \, dx\).

Integrate each term separately:

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

Evaluate from \(-1\) to \(0\):

\(\left[ -\ln|2-x| - \frac{1}{2} \ln|3+2x| - \frac{3}{2(3+2x)} \right]_{-1}^{0} = 1 + \frac{1}{2} \ln \left( \frac{3}{4} \right)\).

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

Assume \(\frac{5x^2 + x + 27}{(2x + 1)(x^2 + 9)} = \frac{A}{2x+1} + \frac{Bx + C}{x^2 + 9}\).

Multiply through by the denominator: \(5x^2 + x + 27 = A(x^2 + 9) + (Bx + C)(2x + 1)\).

Expand and equate coefficients:

\(5x^2 + x + 27 = Ax^2 + 9A + 2Bx^2 + Bx + Cx + C\).

\(5x^2 + x + 27 = (A + 2B)x^2 + (B + C)x + (9A + C)\).

Equating coefficients gives:

\(A + 2B = 5\)

\(B + C = 1\)

\(9A + C = 27\)

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

Thus, \(f(x) = \frac{3}{2x+1} + \frac{x}{x^2 + 9}\).

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

\(\int_0^4 f(x) \, dx = \int_0^4 \left( \frac{3}{2x+1} + \frac{x}{x^2 + 9} \right) \, dx\).

Integrate each term separately:

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

\(\int \frac{x}{x^2 + 9} \, dx = \frac{1}{2} \ln |x^2 + 9|\).

Evaluate from 0 to 4:

\(\left[ \frac{3}{2} \ln |2x+1| \right]_0^4 + \left[ \frac{1}{2} \ln |x^2 + 9| \right]_0^4\).

\(= \frac{3}{2} (\ln 9 - \ln 1) + \frac{1}{2} (\ln 25 - \ln 9)\).

\(= \frac{3}{2} \ln 9 + \frac{1}{2} \ln \frac{25}{9}\).

\(= \ln 9^{3/2} + \ln \left( \frac{25}{9} \right)^{1/2}\).

\(= \ln 27 + \ln \frac{5}{3}\).

\(= \ln (27 \times \frac{5}{3})\).

\(= \ln 45\).

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

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

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

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

Expand and collect terms:

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

Equate coefficients:

\(3A + C = 3\)

\(2A + 3B = 0\)

\(2B = -4\)

Solve for \(B\):

\(B = -2\).

Substitute \(B = -2\) into \(2A + 3B = 0\):

\(2A - 6 = 0\)

\(A = 3\).

Substitute \(A = 3\) into \(3A + C = 3\):

\(9 + C = 3\)

\(C = -6\).

Thus, \(f(x) = \frac{3}{x} - \frac{2}{x^2} - \frac{6}{3x+2}\).

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

\(\int_1^2 \left( \frac{3}{x} - \frac{2}{x^2} - \frac{6}{3x+2} \right) \, dx\).

Integrate each term separately:

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

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

\(\int \frac{-6}{3x+2} \, dx = -2 \ln |3x+2|\)

Evaluate from 1 to 2:

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

\(= \left( 3 \ln 2 + 1 - 2 \ln 8 \right) - \left( 0 + 2 - 2 \ln 5 \right)\)

\(= 3 \ln 2 - 2 \ln 8 - 2 + 2 \ln 5\)

\(= \ln\left(\frac{8}{8}\right) - 1\)

\(= \ln\left(\frac{25}{8}\right) - 1\).

(i) To show that \((x + 1)\) is a factor of \(4x^3 - x^2 - 11x - 6\), substitute \(x = -1\) into the polynomial:

\(4(-1)^3 - (-1)^2 - 11(-1) - 6 = -4 - 1 + 11 - 6 = 0\).

Since the result is 0, \((x + 1)\) is a factor.

(ii) To find \(\int \frac{4x^2 + 9x - 1}{4x^3 - x^2 - 11x - 6} \, dx\), perform polynomial division of \(4x^3 - x^2 - 11x - 6\) by \(x + 1\) to obtain the quotient \(4x^2 - 5x - 6\).

Express the integrand as partial fractions:

\(\frac{4x^2 + 9x - 1}{4x^3 - x^2 - 11x - 6} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{4x+3}\).

Solving for constants, we find \(A = -2\), \(B = 1\), \(C = 8\).

Integrate each term:

\(-2 \int \frac{1}{x+1} \, dx + \int \frac{1}{x-2} \, dx + 2 \int \frac{1}{4x+3} \, dx\).

The integrals yield:

\(-2 \ln|x+1| + \ln|x-2| + 2 \ln|4x+3|\).

Thus, the solution is \(-2 \ln(x+1) + \ln(x-2) + 2 \ln(4x+3)\).

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

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

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

\(11x + 7 = A(x+2)^2 + B(2x-1)(x+2) + C(2x-1)\)

Expand and equate coefficients to find \(A = 2\), \(B = -1\), \(C = 3\).

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

(ii) Integrate each term separately:

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

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

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

Evaluate from 1 to 2:

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

\(= \left( \ln(3) - \ln(4) - \frac{3}{4} \right) - \left( \ln(1) - \ln(3) - \frac{3}{3} \right)\)

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

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

\(= \ln\left(\frac{9}{4}\right) + \frac{1}{4}\)

(i) To express \(f(x)\) in the form \(\frac{A}{2-x} + \frac{Bx+C}{2+x^2}\), we equate:

\(\frac{6 + 6x}{(2-x)(2+x^2)} = \frac{A}{2-x} + \frac{Bx+C}{2+x^2}\)

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

\(6 + 6x = A(2+x^2) + (Bx+C)(2-x)\)

Expand and collect like terms:

\(6 + 6x = 2A + Ax^2 + 2Bx + Cx - Bx^2 - Cx\)

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

Equating coefficients, we get:

\(A - B = 0\)

\(2B + C = 6\)

\(2A = 6\)

Solving these equations, we find:

\(A = 3\)

\(B = 3\)

\(C = 0\)

(ii) To show \(\int_{-1}^{1} f(x) \, dx = 3 \ln 3\), integrate each term separately:

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

\(\int \frac{3x}{2+x^2} \, dx = \frac{3}{2} \ln |2+x^2|\)

\(\int \frac{0}{2+x^2} \, dx = 0\)

Evaluate from \(-1\) to \(1\):

\(-3 \ln |2-1| + \frac{3}{2} \ln |2+1^2| - (-3 \ln |2+1| + \frac{3}{2} \ln |2+(-1)^2|)\)

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

\(= 3 \ln 3\)

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

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

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

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

Expand and equate coefficients to find \(A = 1\), \(B = -2\), \(C = 3\).

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

(b) Integrate \(f(x)\) from 0 to 4:

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

Integrate each term separately:

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

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

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

Evaluate from 0 to 4:

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

Substitute limits:

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

Simplify to obtain \(1 - \ln 3\).

To express \(\frac{7x^2 - 3x + 2}{x(x^2 + 1)}\) in partial fractions, we assume the form:

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

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

\(7x^2 - 3x + 2 = A(x^2 + 1) + (Bx + C)x\)

Expanding the right side gives:

\(7x^2 - 3x + 2 = Ax^2 + A + Bx^2 + Cx\)

Combine like terms:

\(7x^2 - 3x + 2 = (A + B)x^2 + Cx + A\)

Equate coefficients:

1. \(A + B = 7\)

2. \(C = -3\)

3. \(A = 2\)

From equation 3, \(A = 2\).

Substitute \(A = 2\) into equation 1:

\(2 + B = 7\)

\(B = 5\)

Thus, the partial fraction decomposition is:

\(\frac{2}{x} + \frac{5x - 3}{x^2 + 1}\)

(i) Let \(u = \sqrt{6-x}\), then \(u^2 = 6-x\) and differentiating gives \(2u \, du = -dx\), or \(dx = -2u \, du\).

Substitute \(x = 6-u^2\) into the integral:

\(I = \int \frac{5}{6-u^2 + u} (-2u) \, du\).

Simplify the integrand:

\(I = \int \frac{-10u}{6-u^2+u} \, du = \int \frac{-10u}{(3-u)(2+u)} \, du\).

Change the limits: when \(x = 2\), \(u = \sqrt{4} = 2\); when \(x = 5\), \(u = \sqrt{1} = 1\).

Thus, \(I = \int_{1}^{2} \frac{10u}{(3-u)(2+u)} \, du\).

(ii) Use partial fraction decomposition:

\(\frac{10u}{(3-u)(2+u)} = \frac{A}{3-u} + \frac{B}{2+u}\).

Multiply through by \((3-u)(2+u)\) and equate coefficients to find \(A = 6\) and \(B = -4\).

Integrate each term separately:

\(\int \frac{6}{3-u} \, du = -6 \ln|3-u|\) and \(\int \frac{-4}{2+u} \, du = -4 \ln|2+u|\).

Combine and evaluate from 1 to 2:

\(I = [-6 \ln|3-u| - 4 \ln|2+u|]_{1}^{2}\).

Calculate:

\(I = [-6 \ln(1) - 4 \ln(4)] - [-6 \ln(2) - 4 \ln(3)]\).

Simplify to find \(I = 2 \ln\left(\frac{9}{2}\right)\).

(i) Since \((2x - 1)\) is a factor of \(p(x)\), substituting \(x = \frac{1}{2}\) into \(p(x)\) should yield zero:

\(p\left(\frac{1}{2}\right) = a\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - a = 0\)

\(\Rightarrow \frac{a}{8} - \frac{1}{4} + 2 - a = 0\)

\(\Rightarrow \frac{a}{8} - a = -\frac{7}{4}\)

\(\Rightarrow -\frac{7a}{8} = -\frac{7}{4}\)

\(\Rightarrow a = 2\)

Substitute \(a = 2\) back into \(p(x)\):

\(p(x) = 2x^3 - x^2 + 4x - 2\)

Factorise \(p(x)\) using \((2x - 1)\):

\(p(x) = (2x - 1)(x^2 + 2)\)

(ii) Express \(\frac{8x - 13}{p(x)}\) in partial fractions:

\(\frac{8x - 13}{(2x-1)(x^2+2)} = \frac{A}{2x-1} + \frac{Bx + C}{x^2+2}\)

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

\(8x - 13 = A(x^2 + 2) + (Bx + C)(2x - 1)\)

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

\(A = -4\), \(B = 2\), \(C = 5\)

Thus, \(\frac{8x - 13}{p(x)} = \frac{-4}{2x-1} + \frac{2x + 5}{x^2 + 2}\)

(i) To express \(f(x)\) in the form \(\frac{A}{2-x} + \frac{Bx+C}{4+x^2}\), we equate:

\(\frac{12 + 8x - x^2}{(2-x)(4+x^2)} = \frac{A}{2-x} + \frac{Bx+C}{4+x^2}\)

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

\(12 + 8x - x^2 = A(4+x^2) + (Bx+C)(2-x)\)

Expand and collect like terms:

\(12 + 8x - x^2 = 4A + Ax^2 + 2Bx + Cx - Bx^2 - Cx\)

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

Equating coefficients, we get:

\(A - B = -1\)

\(2B + C = 8\)

\(4A = 12\)

Solving these equations gives \(A = 3\), \(B = 4\), \(C = 0\).

(ii) To show \(\int_0^1 f(x) \, dx = \ln\left(\frac{25}{2}\right)\), integrate each term separately:

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

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

\(\int \frac{0}{4+x^2} \, dx = 0\)

Evaluate from 0 to 1:

\(-3 \ln|2-1| + 2 \ln|4+1^2| - (-3 \ln|2-0| + 2 \ln|4+0^2|)\)

\(= -3 \ln(1) + 2 \ln(5) + 3 \ln(2) - 2 \ln(4)\)

\(= 2 \ln(5) + 3 \ln(2) - 2 \ln(4)\)

\(= \ln\left(\frac{5^2 \cdot 2^3}{4^2}\right)\)

\(= \ln\left(\frac{25}{2}\right)\)

First, express \(\frac{2x + 7}{(2x + 1)(x + 2)}\) as partial fractions:

\(\frac{2x + 7}{(2x + 1)(x + 2)} = \frac{A}{2x + 1} + \frac{B}{x + 2}\).

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

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

Expanding gives:

\(2x + 7 = Ax + 2A + 2Bx + B\).

Combine like terms:

\(2x + 7 = (A + 2B)x + (2A + B)\).

Equating coefficients, we get:

\(A + 2B = 2\)

\(2A + B = 7\).

Solving these equations, we find:

\(A = 4\), \(B = -1\).

Thus, \(\frac{2x + 7}{(2x + 1)(x + 2)} = \frac{4}{2x + 1} - \frac{1}{x + 2}\).

Now integrate each term separately:

\(\int \left( \frac{4}{2x + 1} - \frac{1}{x + 2} \right) \, dx = 2 \ln |2x + 1| - \ln |x + 2| + C\).

Apply the limits from 0 to 7:

\(\left[ 2 \ln |2x + 1| - \ln |x + 2| \right]_0^7\).

Calculate at the upper limit (7):

\(2 \ln |15| - \ln |9| = 2 \ln 15 - \ln 9\).

Calculate at the lower limit (0):

\(2 \ln |1| - \ln |2| = 0 - \ln 2\).

Subtract the lower limit from the upper limit:

\((2 \ln 15 - \ln 9) - (0 - \ln 2) = 2 \ln 15 - \ln 9 + \ln 2\).

Simplify using logarithm properties:

\(2 \ln 15 - \ln 9 + \ln 2 = \ln \left( \frac{15^2 \cdot 2}{9} \right) = \ln 50\).

(i) Express \(\frac{2}{(x+1)(x+3)}\) as \(\frac{A}{x+1} + \frac{B}{x+3}\).

Multiply through by \((x+1)(x+3)\) to get: \(2 = A(x+3) + B(x+1)\).

Set \(x = -1\): \(2 = A(-1+3) \Rightarrow A = 1\).

Set \(x = -3\): \(2 = B(-3+1) \Rightarrow B = -1\).

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

(ii) Square the result from part (i):

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

Substitute \(\frac{2}{(x+1)(x+3)} = \frac{1}{x+1} - \frac{1}{x+3}\) to get:

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

(iii) Integrate \(\int_0^1 \frac{4}{(x+1)^2(x+3)^2} \, dx\).

Using the result from part (ii):

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

Integrate each term:

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

Evaluate from 0 to 1:

\(-\frac{1}{2} - \ln 2 + \ln 4 - \frac{1}{4} = \frac{7}{12} - \ln \frac{3}{2}\).

To solve \(\int_1^2 \frac{2}{u(4-u)} \, du\), we use partial fraction decomposition.

Express \(\frac{2}{u(4-u)}\) as \(\frac{A}{u} + \frac{B}{4-u}\).

Equating coefficients, we have:

\(2 = A(4-u) + Bu\)

\(2 = 4A - Au + Bu\)

\(2 = 4A + (B-A)u\)

Comparing coefficients, we get:

\(4A = 2\) and \(B-A = 0\)

Solving these, \(A = \frac{1}{2}\) and \(B = \frac{1}{2}\).

Thus, \(\frac{2}{u(4-u)} = \frac{1/2}{u} + \frac{1/2}{4-u}\).

Integrate each term separately:

\(\int \frac{1/2}{u} \, du = \frac{1}{2} \ln |u|\)

\(\int \frac{1/2}{4-u} \, du = -\frac{1}{2} \ln |4-u|\)

Evaluate from 1 to 2:

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

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

\(= 0 - (-\frac{1}{2} \ln 3)\)

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

(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 \(2 \ln \frac{3}{2} + \frac{1}{\sqrt{3}} \pi\).

(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:

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

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