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