Express $\frac{4x^3-x^2+37x-50}{(x^2+9)(x-1)}$ in partial fractions.

1. Recognise that the fraction is improper and set up the correct partial fraction form

The degree of the numerator (3) is equal to the degree of the denominator (3). Thus, there is a constant term.
Let $\frac{4x^3-x^2+37x-50}{(x^2+9)(x-1)} = A + \frac{Bx+C}{x^2+9} + \frac{D}{x-1}$

2. Find the constant term $A$ by comparing the leading coefficients

The leading term of the numerator is $4x^3$ and the leading term of the denominator is $x^3$.
Therefore, $A = \frac{4x^3}{x^3} = 4$.
So, $\frac{4x^3-x^2+37x-50}{(x^2+9)(x-1)} = 4 + \frac{Bx+C}{x^2+9} + \frac{D}{x-1}$

3. Form the basic identity

$4x^3-x^2+37x-50 = 4(x^2+9)(x-1) + (Bx+C)(x-1) + D(x^2+9)$

4. Substitute $x = 1$ to find $D$

Let $x = 1$:
$4(1)^3 - (1)^2 + 37(1) - 50 = 4(10)(0) + (B+C)(0) + D(1^2+9)$
$4 - 1 + 37 - 50 = 10D$
$-10 = 10D \implies D = -1$

5. Substitute $x = 0$ to find $C$

Let $x = 0$:
$-50 = 4(9)(-1) + C(-1) + D(9)$
$-50 = -36 - C + 9(-1)$
$-50 = -45 - C \implies C = 5$

6. Compare coefficients of $x^2$ to find $B$

Comparing coefficients of $x^2$:
$-1 = 4(-1) + B + D$
$-1 = -4 + B - 1$
$-1 = B - 5 \implies B = 4$

Therefore, the partial fractions are $4 + \frac{4x+5}{x^2+9} - \frac{1}{x-1}$.