(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:
\(\left[ 2 \ln |2x-1| - \frac{1}{2} \ln |x^2+2| \right]_1^5\)
\(= \left( 2 \ln 9 - \frac{1}{2} \ln 27 \right) - \left( 2 \ln 1 - \frac{1}{2} \ln 3 \right)\)
\(= 2 \ln 9 - \frac{1}{2} \ln 27 + \frac{1}{2} \ln 3\)
\(= \ln 81 - \ln 27 + \ln 3\)
\(= \ln \left( \frac{81 \times 3}{27} \right)\)
\(= \ln 27\)
