Answer
$\ln{\frac{3}{8}}$
Work Step by Step
Factor the denominator:
$\int_0^1{\frac{x-4}{x^{2}-5x+6}}dx$ = $\int_0^1{\frac{x-4}{(x-2)(x-3)}}dx$
Write the fraction as a sum of two fractions:
$\frac{x-4}{(x-2)(x-3)}$ = $\frac{A}{x-2}+\frac{B}{x-3}$
Determine $A$ and $B$:
$x-4$ = $A(x-3)+B(x-2)$
$$\begin{align*}
\begin{cases}
A+B &= 1\\
-3A-2B &= -4.
\end{cases}
\end{align*}$$
$A = 2$, $B = -1$
So
$\int_0^1{\frac{x-4}{x^{2}-5x+6}}dx$ = $\int_0^1\left({\frac{2}{x-2}+\frac{-1}{x-3}}\right)dx$
$=[2\ln|x-2|-\ln|x-3|]_0^1$
$=(2\ln1-\ln2)-(2\ln2-\ln3)$ = $-3\ln2+\ln3$
$=\ln{\frac{3}{8}}$