Answer
$$\int_{0}^{1}\frac{x-4}{x^{2}-5x+6}dx=ln\frac{3}{8}$$
Work Step by Step
$$\int_{0}^{1}\frac{x-4}{x^{2}-5x+6}dx=\int_{0}^{1}\frac{(2x-6)-(x-2)}{(x-2)(x-3)}dx$$
$$=\int_{0}^{1}(\frac{2}{x-2}-\frac{1}{x-3})dx$$
$$=\left [2ln\left | x-2 \right |-ln\left | x-3 \right|\right ]_{0}^{1}$$
$$=ln3-3ln2=ln\frac{3}{8}$$
