Resolve the fraction (2x2 + 7x + 33) / (x3 + x2 - 11x) into partial fractions.

Step 1: Factor the denominator.

$x^3 + 2x^2 - 11x = x(x^2 + 2x - 11)$

Step 2: Set up the partial fraction decomposition.

$\frac{2x^2 + 7x + 23}{x(x^2 + 2x - 11)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 2x - 11}$

Step 3: Multiply through by the denominator $x(x^2 + 2x - 11)$.

$2x^2 + 7x + 23 = A(x^2 + 2x - 11) + (Bx + C)x$

Step 4: Expand the right-hand side.

$A(x^2 + 2x - 11) + (Bx + C)x = Ax^2 + 2Ax - 11A + Bx^2 + Cx = (A + B)x^2 + (2A + C)x - 11A$

Step 5: Equate coefficients of corresponding powers of $x$.

Step 6: Solve for $A$.

$A = \frac{23}{-11} = -\frac{23}{11}$

Step 7: Solve for $B$.

$B = 2 - A = 2 - \left(-\frac{23}{11}\right) = 2 + \frac{23}{11} = \frac{22}{11} + \frac{23}{11} = \frac{45}{11}$

Step 8: Solve for $C$.

$C = 7 - 2A = 7 - 2\left(-\frac{23}{11}\right) = 7 + \frac{46}{11} = \frac{77}{11} + \frac{46}{11} = \frac{123}{11}$

Step 9: Substitute $A$, $B$, and $C$ into the decomposition.

$\frac{2x^2 + 7x + 23}{x^3 + 2x^2 - 11x} = \frac{-\frac{23}{11}}{x} + \frac{\frac{45}{11}x + \frac{123}{11}}{x^2 + 2x - 11}$

Step 10: Simplify the expression.

$\frac{2x^2 + 7x + 23}{x^3 + 2x^2 - 11x} = \frac{1}{11}\left( \frac{-23}{x} + \frac{45x + 123}{x^2 + 2x - 11} \right)$

$\boxed{\dfrac{1}{11}\left( \dfrac{-23}{x} + \dfrac{45x + 123}{x^{2} + 2x - 11} \right)}$