Q.Resolve $\dfrac{3x-1}{(x-1)(x-2)}$ into partial fractions.

Both factors are linear and non-repeated, so use the cover-up (substitution) method to find each constant.

Let $\dfrac{5x+6}{(x+2)(1-x)} = \dfrac{A}{x+2}+\dfrac{B}{1-x}$

Multiplying both sides by $(x+2)(1-x)$:

$5x+6 = A(1-x) + B(x+2)$

Put $x=-2$: $5(-2)+6 = A(1-(-2)) \Rightarrow -4 = 3A \Rightarrow A=-\dfrac{4}{3}$

Put $x=1$: $5(1)+6 = B(1+2) \Rightarrow 11 = 3B \Rightarrow B=\dfrac{11}{3}$

$\dfrac{5x+1}{(x+2)(x-1)} = \dfrac{3}{x+2} + \dfrac{2}{x-1}$.

Since the denominator has distinct linear factors, write

$\dfrac{5x+1}{(x+2)(x-1)} = \dfrac{A}{x+2} + \dfrac{B}{x-1}$.

Multiply through by $(x+2)(x-1)$:

$5x + 1 = A(x-1) + B(x+2)$.

Put $x = 1$: $5(1)+1 = B(3) \Rightarrow 6 = 3B \Rightarrow B = 2$.

Put $x = -2$: $5(-2)+1 = A(-3) \Rightarrow -9 = -3A \Rightarrow A = 3$.

Hence

$\dfrac{5x+1}{(x+2)(x-1)} = \dfrac{3}{x+2} + \dfrac{2}{x-1}$.