Answer
$$\int\frac{2x^3-2x^2+1}{x^2-x}dx=x^2-\ln|x|+\ln|x-1|+C$$
Work Step by Step
$$I=\int\frac{2x^3-2x^2+1}{x^2-x}dx$$
1) Perform long division:
We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{2x^3-2x^2+1}{x^2-x}=2x+\frac{1}{x^2-x}$$
2) Express the remainder fraction as a sum of partial fractions:
$$\frac{1}{x^2-x}=\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$$
Clear fractions:
$$Ax-A+Bx=1$$ $$(A+B)x-A=1$$
Equating coefficients of corresponding powers of $x$, we get
$-A=1$, so $A=-1$
$A+B=0$, so $B=1$
Therefore,
$$\frac{1}{x^2-x}=\frac{1}{x(x-1)}=-\frac{1}{x}+\frac{1}{x-1}$$
2) Evaluate the integral:
$$I=\int2xdx-\int\frac{1}{x}dx+\int\frac{1}{x-1}x$$ $$I=x^2-\ln|x|+\ln|x-1|+C$$
