Integration of rational of polynomials
You need the construction called partial fractions (Wiki link). The idea is that you want to write $$ \frac{10x-8}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} $$ so you can bring the RHS to common denominator and compare the numerators, getting 2 equations in 2 unknowns.
Write: $$\frac{10x-8}{(x-1)(x-2)}= {a\over x-1}+{b\over x-2}$$
then you have to find $a$ and $b$. That you do after clearing the denominators:
$$10x-8 = a(x-2)+b(x-1)$$ so you have to solve a system $10 =a+b$ and $-8=-2a-b$