Integral of $\frac{2}{x^3-x^2}$ How can I integrate $\dfrac2{x^3-x^2}$? Can someone please give me some hints? Thanks a lot!

$$\int \frac2{x^3-x^2}\, dx =\frac2{x^2(x-1)}= \int \left(\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}\right)\,dx$$
$$Ax(x-1) + B(x-1) + Cx^2 = 2$$
We choose values for $x$ that "zero out" some of the terms, to simplifying the computation of the unknown values $A, B, C$:
$$x = 1 \implies A(1)(1 - 1) + B(1-1) + C(1^2) = 2 \iff  0 + 0 + C = 2\quad \checkmark$$
$$x = 0 \implies B(-1) = 2 \implies B = -2\quad \checkmark$$
$$x = 2, C = 2, B = -2 \implies 2A - 2 + 8 = 2\iff A = -2\quad \checkmark$$
$$ \int \left(\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}\right)\,dx = \int \left(-\frac{2}x -\frac{2}{x^2}+\frac{2}{x-1}\right)\,dx$$ $$ =-2 \int \frac{dx}x \quad -\quad  2\int \frac{dx}{x^2}\quad + \quad 2 \int \frac{dx}{x-1}$$
