## Thomas' Calculus 13th Edition

$\ln |x-1|+2 \ln |x-2| +c$
The integration by parts formula suggests that $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$ Now, we have $I=\int \dfrac{x}{x^2-3x+2}=\int \dfrac{x}{(x-1)(x-2)}$ Now, need to write the integral into partial fractions. So, $I=\int \dfrac{x}{(x-1)(x-2)}=\int \dfrac{1}{(x-1)} dx +\int \dfrac{2}{(x-2)} dx$ Thus, $I=\int \dfrac{x}{(x-1)(x-2)}=\ln |x-1|+2 \ln |x-2| +c$