Evaluate $\displaystyle \int \dfrac{x^3 - 6 x^2 + 10 x - 2}{x^2 - 5 x + 6} \; dx$
Let $I = \displaystyle \int \dfrac{x^3 - 6 x^2 + 10 x - 2}{x^2 - 5 x + 6} \; dx$ $\;\;\; \cdots \; (1)$
$\begin{array}{rll}
x^2 - 5 x + 6 ) & x^3 - 6 x^2 + 10 x - 2 & (x - 1 \\
& \underline{x^3 - 5 x^2 + 6 x} & \\
& \hspace{6mm} - x^2 \; + 4 x \; - 2 & \\
& \hspace{6mm} \underline{- x^2 \; + 5 x \; - 6} & \\
& \hspace{15mm} - x \; + \; 4 &
\end{array}$
$\therefore$ $\;$ $\dfrac{x^3 - 6 x^2 + 10 x - 2}{x^2 - 5 x + 6} = x - 1 + \dfrac{4 - x}{x^2 - 5 x + 6}$ $\;\;\; \cdots \; (2)$
In view of equation $(2)$, equation $(1)$ becomes
$I = \displaystyle \int x \; dx - \displaystyle \int dx + \displaystyle \int \dfrac{4 - x}{x^2 - 5 x + 6} \; dx$ $\;\;\; \cdots \; (3)$
Now, $\displaystyle \int x \; dx = \dfrac{x^2}{2} + c_1$ $\;\;\; \cdots \; (4)$
$\displaystyle \int dx = x + c_2$ $\;\;\; \cdots (5)$
$\displaystyle \int \dfrac{4 - x}{x^2 - 5 x + 6} \; dx = \displaystyle \int \dfrac{4 - x}{\left(x - 2\right) \left(x - 3\right)} \; dx$ $\;\;\; \cdots \; (6)$
Let $\dfrac{4 - x}{\left(x - 2\right) \left(x - 3\right)} = \dfrac{A}{x - 2} + \dfrac{B}{x - 3}$ $\;\;\; \cdots \; (6a)$
i.e. $4 - x = A \left(x - 3\right) + B \left(x - 2\right)$
When $x = 2$, $2 = - A$ $\implies$ $A = -2$
When $x = 3$, $B = 1$
Substituting the values of A and B in equation $(6a)$ gives
$\dfrac{4 - x}{\left(x - 2\right) \left(x - 3\right)} = \dfrac{-2}{x - 2} + \dfrac{1}{x - 3}$ $\;\;\; \cdots \; (6b)$
$\therefore$ $\;$ In view of equation $(6b)$ equation $(6)$ becomes
$\begin{aligned}
\int \dfrac{4 - x}{x^2 - 5 x + 6} \; dx & = - 2 \int \dfrac{dx}{x - 2} + \int \dfrac{dx}{x - 3} \\\\
& = -2 \log \left|x - 2\right| + \log \left|x - 3\right| + c_3 \;\;\; \cdots \; (6c)
\end{aligned}$
$\therefore$ $\;$ In view of equations $(4)$, $(5)$ and $(6c)$, equation $(3)$ becomes
$I = \dfrac{x^2}{2} - x - 2 \log \left|x - 2\right| + \log \left|x - 3\right| + c$
where $c = c_1 - c_2 + c_3$