Integration by partial-fractions, I´m stuck in this one.

Question

I don´t really know how get the factors in the denominator which allow me to use a case

$\int\frac{x^2+1}{x^2-x} dx$

Answer 1

$$\frac{x^2+1}{x^2-x}=\frac{x^2}{x^2-x}+\frac{1}{x^2-x}$$ $$=\frac{x^2}{x(x-1)}+\frac{1}{x(x-1)}$$ $$=\frac{x}{(x-1)}+\frac{1}{x(x-1)}$$ $$=\frac{x-1+1}{(x-1)}+\frac{1}{x(x-1)}$$ $$=1+\frac{1}{(x-1)}+\frac{1}{x(x-1)}$$ now let $$\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$$

equate the two sides to find $$A=-1$$ $$B=1$$ so $$=1+\frac{1}{(x-1)}+\frac{1}{(x-1)}-\frac{1}{x}$$ $$=1+\frac{2}{(x-1)}-\frac{1}{x}$$