Decompose (6x^2 + x - 37) / (x - 3)(x + 2)(x - 1) into partial fractions A/(x - 3) + B/(x + 2) + C/(x - 1) and solve for C.

$\dfrac{6x^2 + x - 37}{(x - 3)(x + 2)(x - 1)} = \dfrac{A}{x - 3} + \dfrac{B}{x + 2} + \dfrac{C}{x - 1}$

Multiply both sides by $(x - 3)(x + 2)(x - 1)$
$6x^2 + x - 37 = A(x + 2)(x - 1) + B(x - 3)(x - 1) + C(x - 3)(x + 2)$

Set $x = 1$
$6 + 1 - 37 = A(1 + 2)(1 - 1) + B(1 - 3)(1 - 1) + C(1 - 3)(1 + 2)$

$-30 = -6C$

$C = 5$